Solution Manual for Calculus, 8th Edition

2 LIMITS AND DERIVATIVES 2.1 The Tangent and Velocity Problems 1. (a) Using  (15 250), we construct the following table:   slope =   5 (5 694) 694−250 = − 444 = −444 5−15 10 10 (10 444) 444−250 = − 194 = −388 10−15 5 20 (20 111) 111−250 = − 139 = −278 20−15 5 25 (25 28) 28−250 222 25−15 = − 10 = −222 30 (30 0) 0−250 250 30−15 = − 15 = −166 (b) Using the values of  that correspond to the points closest to  ( = 10 and  = 20), we have −388 + (−278) = −333 2 (c) From the graph, we can estimate the slope of the tangent line at  to be −300 = −333. 9 − 2530 2. (a) Slope = 2948 = 418 42 − 36 6 ≈ 6967 − 2661 (b) Slope = 2948 = 287 42 − 38 4 = 7175 − 2948 = 132 (d) Slope = 3080 44 − 42 2 = 66 − 2806 = 142 = 71 (c) Slope = 2948 42 − 40 2 From the data, we see that the patient’s heart rate is decreasing from 71 to 66 heartbeatsminute after 42 minutes. After being stable for a while, the patient’s heart rate is dropping. 3. (a)  = 1 ,  (2 −1) 1−  (i) 15 (ii) 19 (iii) 199 (iv) 1999 (v) 25 (vi) 21 (vii) 201 (viii) 2001 (b) The slope appears to be 1. (c) Using  = 1, an equation of the tangent line to the ( 1(1 − ))   (15 −2) 2 (199 −1010 101) 1010 101 (25 −0666 667) 0666 667 (201 −0990 099) 0990 099 (19 −1111 111) 1111 111 (1999 −1001 001) 1001 001 (21 −0909 091) 0909 091 (2001 −0999 001) 0999 001 curve at  (2 −1) is  − (−1) = 1( − 2), or  =  − 3. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 67 68 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 4. (a)  = cos ,  (05 0)  (b) The slope appears to be −.    (i) 0 (0 1) (ii) 04 (04 0309017) (iii) 049 (049 0031411) (iv) 0499 (0499 0003142) (1 −1) (v) 1 (vi) 06 (vii) 051 (viii) 0501 −2 (06 −0309017) (051 −0031411) (0501 −0003142) (c)  − 0 = −( − 05) or  = − + 12 . (d) −3090170 −3141076 −3141587 −2 −3090170 −3141076 −3141587 5. (a)  = () = 40 − 162 . At  = 2,  = 40(2) − 16(2)2 = 16. The average velocity between times 2 and 2 +  is ave =   40(2 + ) − 16(2 + )2 − 16 −24 − 162 (2 + ) − (2) = = = −24 − 16, if  6= 0. (2 + ) − 2   (i) [2 25]:  = 05, ave = −32 fts (ii) [2 21]:  = 01, ave = −256 fts (iii) [2 205]:  = 005, ave = −248 fts (iv) [2 201]:  = 001, ave = −2416 fts (b) The instantaneous velocity when  = 2 ( approaches 0) is −24 fts. 6. (a)  = () = 10 − 1862 . At  = 1,  = 10(1) − 186(1)2 = 814. The average velocity between times 1 and 1 +  is ave =   10(1 + ) − 186(1 + )2 − 814 628 − 1862 (1 + ) − (1) = = = 628 − 186, if  6= 0. (1 + ) − 1   (i) [1 2]:  = 1, ave = 442 ms (ii) [1 15]:  = 05, ave = 535 ms (iii) [1 11]:  = 01, ave = 6094 ms (iv) [1 101]:  = 001, ave = 62614 ms (v) [1 1001]:  = 0001, ave = 627814 ms (b) The instantaneous velocity when  = 1 ( approaches 0) is 628 ms. 7. (a) (i) On the interval [2 4] , ave = (4) − (2) 792 − 206 = = 293 fts. 4−2 2 (ii) On the interval [3 4] , ave = 792 − 465 (4) − (3) = = 327 fts. 4−3 1 (iii) On the interval [4 5] , ave = 1248 − 792 (5) − (4) = = 456 fts. 5−4 1 (iv) On the interval [4 6] , ave = 1767 − 792 (6) − (4) = = 4875 fts. 6−4 2 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.1 THE TANGENT AND VELOCITY PROBLEMS ¤ (b) Using the points (2 16) and (5 105) from the approximate tangent line, the instantaneous velocity at  = 3 is about 89 105 − 16 = ≈ 297 fts. 5−2 3 8. (a) (i)  = () = 2 sin  + 3 cos . On the interval [1 2], ave = (ii) On the interval [1 11], ave = 3 − (−3) (2) − (1) = = 6 cms. 2−1 1 −3471 − (−3) (11) − (1) ≈ = −471 cms. 11 − 1 01 (iii) On the interval [1 101], ave = −30613 − (−3) (101) − (1) ≈ = −613 cms. 101 − 1 001 (iv) On the interval [1 1001], ave = −300627 − (−3) (1001) − (1) ≈ = −627 cms. 1001 − 1 0001 (b) The instantaneous velocity of the particle when  = 1 appears to be about −63 cms. 9. (a) For the curve  = sin(10) and the point  (1 0):     2  (2 0) 0 05 (05 0) 15 (15 08660) 17321 06 (06 08660) 14 (14 −04339) −10847 07 (07 07818) 08 (08 1) (12 08660) 43301 09 (09 −03420) 13 12 11  (13 −08230) (11 −02817) −27433 −28173   0 −21651 −26061 −5 34202 As  approaches 1, the slopes do not appear to be approaching any particular value. We see that problems with estimation are caused by the frequent (b) oscillations of the graph. The tangent is so steep at  that we need to take -values much closer to 1 in order to get accurate estimates of its slope. (c) If we choose  = 1001, then the point  is (1001 −00314) and   ≈ −313794. If  = 0999, then  is (0999 00314) and   = −314422. The average of these slopes is −314108. So we estimate that the slope of the tangent line at  is about −314. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 69 70 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 2.2 The Limit of a Function 1. As  approaches 2,  () approaches 5. [Or, the values of  () can be made as close to 5 as we like by taking  sufficiently close to 2 (but  6= 2).] Yes, the graph could have a hole at (2 5) and be defined such that  (2) = 3. 2. As  approaches 1 from the left,  () approaches 3; and as  approaches 1 from the right, () approaches 7. No, the limit does not exist because the left- and right-hand limits are different. 3. (a) lim  () = ∞ means that the values of  () can be made arbitrarily large (as large as we please) by taking  →−3 sufficiently close to −3 (but not equal to −3). (b) lim () = −∞ means that the values of  () can be made arbitrarily large negative by taking  sufficiently close to 4 →4+ through values larger than 4. 4. (a) As  approaches 2 from the left, the values of  () approach 3, so lim  () = 3. →2− (b) As  approaches 2 from the right, the values of  () approach 1, so lim () = 1. →2+ (c) lim  () does not exist since the left-hand limit does not equal the right-hand limit. →2 (d) When  = 2,  = 3, so  (2) = 3. (e) As  approaches 4, the values of  () approach 4, so lim () = 4. →4 (f ) There is no value of  () when  = 4, so  (4) does not exist. 5. (a) As  approaches 1, the values of  () approach 2, so lim () = 2. →1 (b) As  approaches 3 from the left, the values of  () approach 1, so lim  () = 1. →3− (c) As  approaches 3 from the right, the values of  () approach 4, so lim () = 4. →3+ (d) lim  () does not exist since the left-hand limit does not equal the right-hand limit. →3 (e) When  = 3,  = 3, so  (3) = 3. 6. (a) () approaches 4 as  approaches −3 from the left, so lim () = 4. →−3− (b) () approaches 4 as  approaches −3 from the right, so lim () = 4. →−3+ (c) lim () = 4 because the limits in part (a) and part (b) are equal. →−3 (d) (−3) is not defined, so it doesn’t exist. (e) () approaches 1 as  approaches 0 from the left, so lim () = 1. →0− (f ) () approaches −1 as  approaches 0 from the right, so lim () = −1. →0+ (g) lim () does not exist because the limits in part (e) and part (f ) are not equal. →0 (h) (0) = 1 since the point (0 1) is on the graph of . (i) Since lim () = 2 and lim () = 2, we have lim () = 2. →2− →2+ →2 (j) (2) is not defined, so it doesn’t exist. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.2 THE LIMIT OF A FUNCTION ¤ (k) () approaches 3 as  approaches 5 from the right, so lim () = 3. →5+ (l) () does not approach any one number as  approaches 5 from the left, so lim () does not exist. →5− 7. (a) lim () = −1 (b) lim () = −2 →0− →0+ (c) lim () does not exist because the limits in part (a) and part (b) are not equal. →0 (e) lim () = 0 (d) lim () = 2 →2− →2+ (f ) lim () does not exist because the limits in part (d) and part (e) are not equal. →2 (g) (2) = 1 (h) lim () = 3 →4 8. (a) lim () = ∞ (b) lim () = −∞ (c) lim () = ∞ (d) lim () = −∞ →−3 →2− →−1 →2+ (e) The equations of the vertical asymptotes are  = −3,  = −1 and  = 2. 9. (a) lim  () = −∞ (b) lim  () = ∞ (d) lim  () = −∞ (e) lim  () = ∞ →−7 →−3 →6− (c) lim  () = ∞ →0 →6+ (f ) The equations of the vertical asymptotes are  = −7,  = −3,  = 0, and  = 6. 10. lim  () = 150 mg and lim  () = 300 mg. These limits show that there is an abrupt change in the amount of drug in →12− + →12 the patient’s bloodstream at  = 12 h. The left-hand limit represents the amount of the drug just before the fourth injection. The right-hand limit represents the amount of the drug just after the fourth injection. 11. From the graph of  1 +  if   −1    () = 2 if −1 ≤   1 ,   2 −  if  ≥ 1 we see that lim () exists for all  except  = −1. Notice that the → right and left limits are different at  = −1. 12. From the graph of  1 + sin  if   0   if 0 ≤  ≤  ,  () = cos    sin  if    we see that lim () exists for all  except  = . Notice that the → right and left limits are different at  = . c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 71 72 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 13. (a) lim  () = 1 →0− (b) lim () = 0 →0+ (c) lim  () does not exist because the limits →0 in part (a) and part (b) are not equal. 14. (a) lim  () = −1 →0− (b) lim () = 1 →0+ (c) lim  () does not exist because the limits →0 in part (a) and part (b) are not equal. 15. lim  () = −1, lim () = 2,  (0) = 1 →0− →0+ 16. lim  () = 1, lim  () = −2, lim  () = 2, →0 →3− →3+  (0) = −1,  (3) = 1 17. lim  () = 4, →3+ lim  () = 2, lim () = 2, →−2 →3−  (3) = 3, (−2) = 1 19. For  () = 2 − 3 : 2 − 9   () 18. lim  () = 2, lim  () = 0, lim  () = 3, →0− →0+ →4− lim  () = 0,  (0) = 2,  (4) = 1 →4+   () 31 0508 197 29 0491 525 305 0504 132 295 0495 798 301 0500 832 299 0499 165 3001 0500 083 2999 0499 917 30001 0500 008 29999 0499 992 1 2 − 3 = . →3 2 − 9 2 It appears that lim c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.2 20. For  () =  −25   () −5 −35 7 −59 −305 −29 −29 −299 −299 −2999 −29999 −2999 −29,999 21. For  () = 5 − 1 :    () 05 22364 988 01 6487 213 001 5127 110 0001 5012 521 00001 5001 250 It appears that lim →0 −31 31 −301 301 61 −3001 3001 −30001 30,001 It appears that lim  () = −∞ and that →−3+ lim  () = ∞, so lim ln  − ln 4 : −4   () →−3 →−3− 22. For  () =   ()  2 − 3 does not exist. 2 − 9 (2 + )5 − 32 :   () −05 1835 830 05 131312 500 −01 3934 693 01 88410 100 −001 4877 058 001 80804 010 −0001 4987 521 0001 80080 040 −00001 4998 750 00001 80008 000 5 − 1 = 5.  23. For  () = It appears that lim →0    () −05 48812 500 −001 79203 990 −00001 79992 000 −01 72390 100 −0001 79920 040 (2 + )5 − 32 = 80.   () 39 0253 178 41 0246 926 399 0250 313 401 0249 688 3999 0250 031 4001 0249 969 39999 0250 003 40001 0249 997 It appears that lim  () = 025. The graph confirms that result. →4 24. For  () =  ¤ 2 − 3 : 2 − 9  () −295 THE LIMIT OF A FUNCTION 1 + 9 : 1 + 15  () −11 0427 397 −1001 0598 200 −101 0582 008 −10001 0599 820   () −09 0771 405 −0999 0601 800 −099 0617 992 −09999 0600 180 It appears that lim  () = 06. The graph confirms that result. →−1 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 73 74 ¤ CHAPTER 2 25. For  () = LIMITS AND DERIVATIVES sin 3 : tan 2   () sin 3 ±01 1457 847 It appears that lim ±001 1499 575 The graph confirms that result. ±0001 1499 996 ±00001 1500 000 26. For  () = 5 − 1 :   () 01 1746 189 001 1622 459 0001 1610 734 00001 1609 567 →0 tan 2  = 15. () −01 1486 601 −0001 1608 143 −001 1596 556 −00001 1609 308 It appears that lim  () ≈ 16094. The graph confirms that result. →0 27. For  () =  :   () 01 0794 328 001 0954 993 0001 0993 116 00001 0999 079 It appears that lim () = 1. →0+ The graph confirms that result. 28. For  () = 2 ln :  01 001 0001 00001 () It appears that lim  () = 0. −0023 026 →0+ −0000 461 The graph confirms that result. −0000 007 −0000 000 29. (a) From the graphs, it seems that lim →0 cos 2 − cos  = −15. 2 (b)   () ±01 −1493 759 ±0001 −1499 999 ±001 ±00001 −1499 938 −1500 000 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.2 30. (a) From the graphs, it seems that lim sin  →0 sin  THE LIMIT OF A FUNCTION ¤ 75 (b) = 032.   () ±01 0323 068 ±0001 ±00001 0318 310 0318 310 ±001 0318 357 Later we will be able to show that the exact value is 31. lim +1 32. lim +1 33. lim 2− →5+  − 5 →5−  − 5 1 .  = ∞ since the numerator is positive and the denominator approaches 0 from the positive side as  → 5+ . = −∞ since the numerator is positive and the denominator approaches 0 from the negative side as  → 5− . →1 ( − 1)2 = ∞ since the numerator is positive and the denominator approaches 0 through positive values as  → 1. √  = −∞ since the numerator is positive and the denominator approaches 0 from the negative side as  → 3− . →3− ( − 3)5 34. lim 35. Let  = 2 − 9. Then as  → 3+ ,  → 0+ , and lim ln(2 − 9) = lim ln  = −∞ by (5). →3+ →0+ 36. lim ln(sin ) = −∞ since sin  → 0+ as  → 0+ . →0+ 37. lim 1 →(2)+  sec  = −∞ since 38. lim cot  = lim cos  → − sin  → − 1 is positive and sec  → −∞ as  → (2)+ .  = −∞ since the numerator is negative and the denominator approaches 0 through positive values as  → − . 39. lim  csc  = lim →2 −  →2 − sin  = −∞ since the numerator is positive and the denominator approaches 0 through negative values as  → 2− . 2 − 2 ( − 2)  = lim = −∞ since the numerator is positive and the denominator = lim 2 2 →2−  − 4 + 4 →2− ( − 2) →2−  − 2 40. lim approaches 0 through negative values as  → 2− . 41. lim →2+ 2 − 2 − 8 ( − 4)( + 2) = lim = ∞ since the numerator is negative and the denominator approaches 0 through 2 − 5 + 6 →2+ ( − 3)( − 2) negative values as  → 2+ . 42. lim →0+   1 1 − ln  = ∞ since → ∞ and ln  → −∞ as  → 0+ .   43. lim (ln 2 − −2 ) = −∞ since ln 2 → −∞ and −2 → ∞ as  → 0. →0 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 76 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 44. (a) The denominator of  = 2 + 1 2 + 1 is equal to zero when = 2 3 − 2 (3 − 2) (b)  = 0 and  = 32 (and the numerator is not), so  = 0 and  = 15 are vertical asymptotes of the function. 45. (a)  () = 1 . 3 − 1 From these calculations, it seems that lim  () = −∞ and lim  () = ∞. →1− →1+  05 09 099 0999 09999 099999  () −114 −369 −337 −3337 −33337 −33,3337  15 11 101 1001 10001 100001  () 042 302 330 3330 33330 33,3333 (b) If  is slightly smaller than 1, then 3 − 1 will be a negative number close to 0, and the reciprocal of 3 − 1, that is,  (), will be a negative number with large absolute value. So lim () = −∞. →1− If  is slightly larger than 1, then  − 1 will be a small positive number, and its reciprocal,  (), will be a large positive 3 number. So lim  () = ∞. →1+ (c) It appears from the graph of  that lim  () = −∞ and lim  () = ∞. →1− →1+ 46. (a) From the graphs, it seems that lim →0 1 47. (a) Let () = (1 + )  −0001 −00001 −000001 −0000001 0000001 000001 00001 0001 . () 271964 271842 271830 271828 271828 271827 271815 271692 tan 4 = 4.  (b)  ±01 ±001 ±0001 ±00001  () 4227 932 4002 135 4000 021 4000 000 (b) It appears that lim (1 + )1 ≈ 271828, which is approximately . →0 In Section 3.6 we will see that the value of the limit is exactly . c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.2 THE LIMIT OF A FUNCTION ¤ 77 48. (a) No, because the calculator-produced graph of  () =  + ln | − 4| looks like an exponential function, but the graph of  has an infinite discontinuity at  = 4. A second graph, obtained by increasing the numpoints option in Maple, begins to reveal the discontinuity at  = 4. (b) There isn’t a single graph that shows all the features of  . Several graphs are needed since  looks like ln | − 4| for large negative values of  and like  for   5, but yet has the infinite discontiuity at  = 4. A hand-drawn graph, though distorted, might be better at revealing the main features of this function. 49. For  () = 2 − (21000): (a)  1 08 () 0998 000 0638 259 06 04 02 01 005 0358 484 0158 680 0038 851 0008 928 0001 465 (b)  004 002 001 0005 0003 0001  () 0000 572 −0000 614 −0000 907 −0000 978 −0000 993 −0001 000 It appears that lim () = −0001. →0 It appears that lim  () = 0. →0 50. For () = tan  −  : 3 (a)  10 05 01 005 001 0005 () 0557 407 73 0370 419 92 0334 672 09 0333 667 00 0333 346 67 0333 336 67 (b) It seems that lim () = 13 . →0 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 78 ¤ CHAPTER 2 LIMITS AND DERIVATIVES (c) Here the values will vary from one  () 0001 00005 00001 000005 000001 0000001 0333 333 50 0333 333 44 0333 330 00 0333 336 00 0333 000 00 0000 000 00 calculator to another. Every calculator will eventually give false values. (d) As in part (c), when we take a small enough viewing rectangle we get incorrect output. 51. No matter how many times we zoom in toward the origin, the graphs of  () = sin() appear to consist of almost-vertical lines. This indicates more and more frequent oscillations as  → 0. 52. (a) For any positive integer , if  = 1 1 , then  () = tan = tan() = 0. (Remember that the tangent function has   period .) c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° (b) For any nonnegative number , if  =  () = tan SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS     = tan = 1 = tan  + 4 4 ¤ 79 4 , then (4 + 1) 1 (4 + 1) = tan = tan  4 4  + 4 4  (c) From part (a),  () = 0 infinitely often as  → 0. From part (b),  () = 1 infinitely often as  → 0. Thus, lim tan →0 1  does not exist since  () does not get close to a fixed number as  → 0. There appear to be vertical asymptotes of the curve  = tan(2 sin ) at  ≈ ±090 53. and  ≈ ±224. To find the exact equations of these asymptotes, we note that the graph of the tangent function has vertical asymptotes at  = 2 + . Thus, we must have 2 sin  = 2 + , or equivalently, sin  = 4 + 2 . Since −1 ≤ sin  ≤ 1, we must have sin  = ± 4 and so  = ± sin−1 4 (corresponding to  ≈ ±090). Just as 150◦ is the reference angle for 30◦ ,  − sin−1 4 is the   reference angle for sin−1 4 . So  = ±  − sin−1 4 are also equations of vertical asymptotes (corresponding to  ≈ ±224).  0 . As  → − , 1 −  22 → 0+ , and  → ∞. 1 − 22 54. lim  = lim  →− →− 3 − 1 . −1 55. (a) Let  = √ From the table and the graph, we guess that the limit of  as  approaches 1 is 6.   099 0999 09999 101 1001 10001 5925 31 5992 50 5999 25 6075 31 6007 50 6000 75 3 − 1  65. From the graph we obtain the approximate points of intersection  (09314 55) (b) We need to have 55  √ −1 and (10649 65). Now 1 − 09314 = 00686 and 10649 − 1 = 00649, so by requiring that  be within 00649 of 1, we ensure that  is within 05 of 6. 2.3 Calculating Limits Using the Limit Laws 1. (a) lim [ () + 5()] = lim  () + lim [5()] [Limit Law 1] = lim  () + 5 lim () [Limit Law 3] →2 →2 →2 →2 →2 (b) lim [()]3 = →2  3 lim () →2 [Limit Law 6] = ( −2)3 = −8 = 4 + 5(−2) = −6 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 80 ¤ CHAPTER 2 (c) lim →2 LIMITS AND DERIVATIVES    () = lim  () →2 = lim [3 ()] 3 () →2 = →2 () lim () (d) lim [Limit Law 11] [Limit Law 5] →2 √ 4=2 3 lim  () →2 = [Limit Law 3] lim () →2 = 3(4) = −6 −2 (e) Because the limit of the denominator is 0, we can’t use Limit Law 5. The given limit, lim () →2 () , does not exist because the denominator approaches 0 while the numerator approaches a nonzero number. lim [() ()] () () →2 = →2  () lim  () (f) lim [Limit Law 5] →2 = lim () · lim () →2 →2 lim  () [Limit Law 4] →2 = −2 · 0 =0 4 2. (a) lim [ () + ()] = lim  () + lim () →2 →2 →2 [Limit Law 1] = −1 + 2 =1 (b) lim  () exists, but lim () does not exist, so we cannot apply Limit Law 2 to lim [ () − ()]. →0 →0 →0 The limit does not exist. (c) lim [() ()] = lim  () · lim () →−1 →−1 →−1 [Limit Law 4] =1·2 =2 (d) lim  () = 1, but lim () = 0, so we cannot apply Limit Law 5 to lim →3 Note: lim () →3− ()  () = ∞ since () → 0+ as  → 3− and lim →3+ () Therefore, the limit does not exist, even as an infinite limit.   (e) lim 2  () = lim 2 · lim  () [Limit Law 4] →2 →2  () →3 () →3 . The limit does not exist. = −∞ since () → 0− as  → 3+ . (f)  (−1) + lim () is undefined since  (−1) is →2 →−1 = 22 · (−1) not defined. = −4 3. lim (53 − 32 +  − 6) = lim (53 ) − lim (32 ) + lim  − lim 6 →3 →3 →3 →3 →3 [Limit Laws 2 and 1] = 5 lim 3 − 3 lim 2 + lim  − lim 6 [3] = 5(33 ) − 3(32 ) + 3 − 6 [9, 8, and 7] →3 →3 →3 →3 = 105 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.3 4. CALCULATING LIMITS USING THE LIMIT LAWS [Limit Law 4] lim (4 − 3)(2 + 5 + 3) = lim (4 − 3) lim (2 + 5 + 3) →−1 →−1 = = →−1   lim 2 + lim 5 + lim 3 lim 4 − lim 3  →−1 →−1 →−1 →−1 →−1   lim 2 + 5 lim  + lim 3 lim 4 − 3 lim   →−1 →−1 →−1 →−1 →−1 [2, 1] [3] [9, 8, and 7] = (1 + 3)(1 − 5 + 3) = 4(−1) = −4 lim (4 − 2) 4 − 2 →−2 = →−2 22 − 3 + 2 lim (22 − 3 + 2) [Limit Law 5] 5. lim →−2 = = lim 4 − lim 2 →−2 →−2 2 lim 2 − 3 lim  + lim 2 →−2 →−2 →−2 [1, 2, and 3] 16 − 2 2(4) − 3(−2) + 2 [9, 7, and 8] 7 14 = 16 8  √ 6. lim 4 + 3 + 6 = lim (4 + 3 + 6) = →−2 [11] →−2 =  [1, 2, and 3] lim 4 + 3 lim  + lim 6 →−2 →−2 →−2  (−2)4 + 3 (−2) + 6 √ √ = 16 − 6 + 6 = 16 = 4 [9, 8, and 7] = √ √ [Limit Law 4] 7. lim (1 + 3  ) (2 − 62 + 3 ) = lim (1 + 3  ) · lim (2 − 62 + 3 ) →8 →8 =  →8  √   lim 1 + lim 3  · lim 2 − 6 lim 2 + lim 3 →8 →8 →8 →8 √     = 1 + 3 8 · 2 − 6 · 82 + 83 →8 [1, 2, and 3] [7, 10, 9] = (3)(130) = 390 8. lim  →2 2 − 2 3  − 3 + 5 2 2 2 − 2 →2 3 − 3 + 5  2 lim (2 − 2) →2  = lim (3 − 3 + 5) =  [Limit Law 6] lim [5] →2  lim 2 − lim 2 2 →2 →2  = lim 3 − 3 lim  + lim 5 →2 →2 →2  4−2 = 8 − 3(2) + 5  2 4 2 = = 7 49 2 [1, 2, and 3] [9, 7, and 8] c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° ¤ 81 82 ¤ 9. lim →2 CHAPTER 2 LIMITS AND DERIVATIVES   22 + 1 = 3 − 2 22 + 1 →2 3 − 2   lim (22 + 1)  →2 = lim (3 − 2) [Limit Law 11] lim [5] →2   2 lim 2 + lim 1  →2 →2 = 3 lim  − lim 2 [1, 2, and 3] 2(2)2 + 1 = 3(2) − 2 [9, 8, and 7] →2 =  →2  3 9 = 4 2 10. (a) The left-hand side of the equation is not defined for  = 2, but the right-hand side is. (b) Since the equation holds for all  6= 2, it follows that both sides of the equation approach the same limit as  → 2, just as in Example 3. Remember that in finding lim  (), we never consider  = . → 11. lim →5 2 − 6 + 5 ( − 5)( − 1) = lim = lim ( − 1) = 5 − 1 = 4 →5 →5 −5 −5 −3 3 2 + 3 ( + 3)  = lim = lim = = →−3 2 −  − 12 →−3 ( − 4)( + 3) →−3  − 4 −3 − 4 7 12. lim 13. lim →5 2 − 5 + 6 does not exist since  − 5 → 0, but 2 − 5 + 6 → 6 as  → 5. −5 2 + 3 ( + 3)   = lim = lim . The last limit does not exist since lim = −∞ and →4 2 −  − 12 →4 ( − 4)( + 3) →4  − 4 →4−  − 4 14. lim lim  →4+  − 4 = ∞. ( + 3)( − 3) −3 − 3 −6 6 2 − 9 −3 = lim = lim = = = →−3 22 + 7 + 3 →−3 (2 + 1)( + 3) →−3 2 + 1 2(−3) + 1 −5 5 15. lim 16. lim →−1 2(−1) + 1 −1 1 22 + 3 + 1 (2 + 1)( + 1) 2 + 1 = lim = lim = = = →−1 ( − 3)( + 1) →−1  − 3 2 − 2 − 3 −1 − 3 −4 4 (−5 + )2 − 25 (25 − 10 + 2 ) − 25 −10 + 2 (−10 + ) = lim = lim = lim = lim (−10 + ) = −10 →0 →0 →0 →0 →0     17. lim   8 + 12 + 62 + 3 − 8 (2 + )3 − 8 12 + 62 + 3 = lim = lim →0 →0 →0      2 = lim 12 + 6 +  = 12 + 0 + 0 = 12 18. lim →0 19. By the formula for the sum of cubes, we have lim +2 →−2 3 + 8 = lim +2 →−2 ( + 2)(2 − 2 + 4) = lim 1 →−2 2 − 2 + 4 = 1 1 = . 4+4+4 12 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS 20. We use the difference of squares in the numerator and the difference of cubes in the denominator. lim 4 − 1 →1 3 − 1 (2 − 1)(2 + 1) = lim →1 ( − 1)(2 +  + 1) = lim →1 2(2) 4 ( − 1)( + 1)(2 + 1) ( + 1)(2 + 1) = lim = = 2 →1 ( − 1)( +  + 1) 2 +  + 1 3 3 √ 2 √ √ √ 9 +  − 32 9+−3 9+−3 9++3 (9 + ) − 9  = lim √  = lim √ = lim ·√ 21. lim →0 →0 →0    9 +  + 3 →0  9 +  + 3 9++3 = lim →0  1 1  1 √  = lim √ = = →0 3+3 6 9++3 9++3 √ 2 √ √ √ 4 + 1 − 32 4 + 1 − 3 4 + 1 − 3 4 + 1 + 3 √  = lim = lim ·√ 22. lim →2 →2 −2 −2 4 + 1 + 3 →2 ( − 2) 4 + 1 + 3 4 + 1 − 9 4( − 2) √  = lim √  →2 ( − 2) 4 + 1 + 3 4 + 1 + 3 = lim →2 ( − 2) 4 4 2 = lim √ = √ = →2 3 4 + 1 + 3 9+3 1 1 1 1 − − 3 1 3− −1  3  3 = lim · = lim = lim =− 23. lim →3  − 3 →3  − 3 3 →3 3( − 3) →3 3 9 1 1 − (3 + )−1 − 3−1 3 − (3 + ) − 3 +  3 = lim = lim = lim 24. lim →0 →0 →0 (3 + )3 →0 (3 + )3     1 1 1 1 =− =− =− = lim − →0 3(3 + ) lim [3(3 + )] 3(3 + 0) 9 →0 √ 2 √ 2 √ √ √ √ √ √ 1+ − 1− 1+− 1− 1+− 1− 1++ 1−  √ √ = lim √ = lim ·√ 25. lim →0 →0   1 +  + 1 −  →0  1 +  + 1 −  (1 + ) − (1 − ) 2 2  = lim √  = lim √ √ √ √ = lim √ →0  →0  →0 1++ 1− 1++ 1− 1++ 1− 2 2 √ = =1 = √ 2 1+ 1 26. lim →0  1 1 − 2   +  = lim →0  1 1 −  ( + 1)  = lim →0 1 +1−1 1 = lim = =1 →0  + 1 ( + 1) 0+1 √ √ √ 4−  (4 −  )(4 +  ) 16 −  √ √ = lim = lim →16 16 − 2 →16 (16 − 2 )(4 +  ) →16 (16 − )(4 +  ) 27. lim = lim 1 →16 (4 + 1 1 1 √  = √ = =  16(8) 128 ) 16 4 + 16 2 − 4 + 4 ( − 2)2 ( − 2)2 = lim = lim →2 4 − 32 − 4 →2 (2 − 4)(2 + 1) →2 ( + 2)( − 2)(2 + 1) 28. lim = lim −2 →2 ( + 2)(2 + 1) = 0 =0 4·5 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° ¤ 83 84 ¤ CHAPTER 2 29. lim →0  LIMITS AND DERIVATIVES 1 1 √ −   1+     √ √ √ 1− 1+ 1+ 1+ 1− 1+ −   = lim √   √ √ √ √ = lim →0  →0 →0  1+  +1 1+ 1+ 1+ 1+ 1+ = lim −1 −1 1   = √   =− √ √ = lim √ →0 2 1+ 1+ 1+ 1+0 1+ 1+0 30. √ √  √ 2 + 9 − 5 2 + 9 + 5 2 + 9 − 5 (2 + 9) − 25 √  √  = lim = lim 2 →−4 →−4 →−4 ( + 4) +4 ( + 4)  + 9 + 5 2 + 9 + 5 lim 2 − 16 ( + 4)( − 4) √  = lim √  →−4 ( + 4) →−4 ( + 4) 2 + 9 + 5 2 + 9 + 5 = lim −8 4 −4 − 4 −4 = =− = √ = lim √ →−4 5+5 5 16 + 9 + 5 2 + 9 + 5 ( + )3 − 3 (3 + 32  + 32 + 3 ) − 3 32  + 32 + 3 = lim = lim →0 →0 →0    31. lim = lim →0 (32 + 3 + 2 ) = lim (32 + 3 + 2 ) = 32 →0  2 − ( + )2 1 1 − 2 2 ( + )  ( + )2 2 2 − (2 + 2 + 2 ) −(2 + ) = lim = lim 32. lim = lim →0 →0 →0 →0 2 ( + )2   2 ( + )2 = lim −(2 + ) →0 2 ( + )2 = 33. (a) −2 2 =− 3 2 · 2  (b) 2  ≈ lim √ 3 1 + 3 − 1 →0 (c) lim →0    () −0001 −0000 1 −0000 01 −0000 001 0000 001 0000 01 0000 1 0001 0666 166 3 0666 616 7 0666 661 7 0666 666 2 0666 667 2 0666 671 7 0666 716 7 0667 166 3 The limit appears to be √  √  √   1 + 3 + 1  1 + 3 + 1  1 + 3 + 1 √ ·√ = lim = lim →0 →0 (1 + 3) − 1 3 1 + 3 − 1 1 + 3 + 1 √  1 1 + 3 + 1 lim 3 →0   1  lim (1 + 3) + lim 1 = →0 →0 3   1  = lim 1 + 3 lim  + 1 →0 →0 3 = = =  1 √ 1+3·0+1 3 [Limit Law 3] [1 and 11] [1, 3, and 7] [7 and 8] 2 1 (1 + 1) = 3 3 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 2 . 3 SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ 85 (b) 34. (a) √ √ 3+− 3 ≈ 029 →0  lim   () −0001 −0000 1 −0000 01 −0000 001 0000 001 0000 01 0000 1 0001 0288 699 2 0288 677 5 0288 675 4 0288 675 2 0288 675 1 0288 674 9 0288 672 7 0288 651 1 The limit appears to be approximately 02887. √ √ √  √ (3 + ) − 3 3+− 3 3++ 3 1 √ √  = lim √ √ = lim √ (c) lim ·√ →0 →0  →0  3++ 3 3++ 3 3++ 3 = lim 1 →0 √ √ lim 3 +  + lim 3 →0 =  [Limit Laws 5 and 1] →0 1 lim (3 + ) + →0 √ 3 1 √ = √ 3+0+ 3 1 = √ 2 3 [7 and 11] [1, 7, and 8] 35. Let () = −2 , () = 2 cos 20 and () = 2 . Then −1 ≤ cos 20 ≤ 1 ⇒ −2 ≤ 2 cos 20 ≤ 2 ⇒  () ≤ () ≤ (). So since lim  () = lim () = 0, by the Squeeze Theorem we have →0 →0 lim () = 0. →0 √ √ √ 3 + 2 sin(), and () = 3 + 2 . Then √ √ √ −1 ≤ sin() ≤ 1 ⇒ − 3 + 2 ≤ 3 + 2 sin() ≤ 3 + 2 ⇒ 36. Let () = − 3 + 2 , () =  () ≤ () ≤ (). So since lim  () = lim () = 0, by the Squeeze Theorem →0 →0 we have lim () = 0. →0   37. We have lim (4 − 9) = 4(4) − 9 = 7 and lim 2 − 4 + 7 = 42 − 4(4) + 7 = 7. Since 4 − 9 ≤  () ≤ 2 − 4 + 7 →4 →4 for  ≥ 0, lim  () = 7 by the Squeeze Theorem. →4 38. We have lim (2) = 2(1) = 2 and lim (4 − 2 + 2) = 14 − 12 + 2 = 2. Since 2 ≤ () ≤ 4 − 2 + 2 for all , →1 →1 lim () = 2 by the Squeeze Theorem. →1 39. −1 ≤ cos(2) ≤ 1   ⇒ −4 ≤ 4 cos(2) ≤ 4 . Since lim −4 = 0 and lim 4 = 0, we have →0 →0   lim 4 cos(2) = 0 by the Squeeze Theorem. →0 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 86 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 40. −1 ≤ sin() ≤ 1 ⇒ −1 ≤ sin() ≤ 1 √ √ √ √ / ≤  sin() ≤  . Since lim ( /) = 0 and ⇒ →0+ √  √  sin() = 0 by the Squeeze Theorem. lim (  ) = 0, we have lim →0+ →0+ 41. | − 3| =  if  − 3 ≥ 0 −3 if  − 3  0 −( − 3)  = if  ≥ 3 −3 if   3 3− Thus, lim (2 + | − 3|) = lim (2 +  − 3) = lim (3 − 3) = 3(3) − 3 = 6 and →3+ →3+ →3+ lim (2 + | − 3|) = lim (2 + 3 − ) = lim ( + 3) = 3 + 3 = 6. Since the left and right limits are equal, →3− →3− →3− lim (2 + | − 3|) = 6. →3 42. | + 6| =  +6 −( + 6) if  + 6 ≥ 0 if  + 6  0  = if  ≥ −6 +6 if   −6 −( + 6) We’ll look at the one-sided limits. lim →−6+ 2 + 12 2( + 6) = lim = 2 and | + 6| +6 →−6+ The left and right limits are different, so lim →−6       lim →−6− 2 + 12 2( + 6) = lim = −2 | + 6| →−6− −( + 6) 2 + 12 does not exist. | + 6| 43. 23 − 2  = 2 (2 − 1) = 2  · |2 − 1| = 2 |2 − 1| |2 − 1| =  2 − 1 −(2 − 1) if 2 − 1 ≥ 0 if 2 − 1  0 =    So 23 − 2  = 2 [−(2 − 1)] for   05. Thus, lim 2 − 1 3 2 →05− |2 −  | = 2 − 1 lim if  ≥ 05 2 − 1 if   05 −(2 − 1) 2 →05−  [−(2 − 1)] = lim →05− −1 −1 −1 = −4. = = 2 (05)2 025 44. Since || = − for   0, we have lim 2 − || 2 − (−) 2+ = lim = lim = lim 1 = 1. →−2 →−2 2 +  →−2 2+ 2+ 45. Since || = − for   0, we have lim  →−2 →0− 1 1 −  ||  = lim →0−  1 1 −  −  = lim 2 →0−  , which does not exist since the denominator approaches 0 and the numerator does not. 46. Since || =  for   0, we have lim →0+ 47. (a)  1 1 −  ||  = lim →0+  1 1 −    = lim 0 = 0. →0+ (b) (i) Since sgn  = 1 for   0, lim sgn  = lim 1 = 1. →0+ →0+ (ii) Since sgn  = −1 for   0, lim sgn  = lim −1 = −1. →0− →0− (iii) Since lim sgn  6= lim sgn , lim sgn  does not exist. →0− →0 →0+ (iv) Since |sgn | = 1 for  6= 0, lim |sgn | = lim 1 = 1. →0 →0 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS  −1 if sin   0   48. (a) () = sgn(sin ) = 0 if sin  = 0   1 if sin   0 (i) lim () = lim sgn(sin ) = 1 since sin  is positive for small positive values of . →0+ →0+ (ii) lim () = lim sgn(sin ) = −1 since sin  is negative for small negative values of . →0− →0− (iii) lim () does not exist since lim () 6= lim (). →0 →0− →0+ (iv) lim () = lim sgn(sin ) = −1 since sin  is negative for values of  slightly greater than . → + → + (v) lim () = lim sgn(sin ) = 1 since sin  is positive for values of  slightly less than . → − → − (vi) lim () does not exist since lim () 6= lim (). → → − → + (b) The sine function changes sign at every integer multiple of , so the (c) signum function equals 1 on one side and −1 on the other side of ,  an integer. Thus, lim () does not exist for  = ,  an integer. → 2 +  − 6 ( + 3)( − 2) = lim | − 2| | − 2| →2+ →2+ 49. (a) (i) lim () = lim →2+ = lim →2+ ( + 3)( − 2) −2 [since  − 2  0 if  → 2+ ] = lim ( + 3) = 5 →2+ (ii) The solution is similar to the solution in part (i), but now | − 2| = 2 −  since  − 2  0 if  → 2− . Thus, lim () = lim −( + 3) = −5. →2− →2− (b) Since the right-hand and left-hand limits of  at  = 2 (c) are not equal, lim () does not exist. →2  2  +1 50. (a)  () = ( − 2)2 if   1 if  ≥ 1 lim  () = lim (2 + 1) = 12 + 1 = 2, →1− →1− lim  () = lim ( − 2)2 = (−1)2 = 1 →1+ (b) Since the right-hand and left-hand limits of  at  = 1 →1+ (c) are not equal, lim () does not exist. →1 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° ¤ 87 88 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 51. For the lim () to exist, the one-sided limits at  = 2 must be equal. lim () = lim →2 →2− lim () = lim →2+ →2+ →2−   4 − 12  = 4 − 1 = 3 and √ √ √  +  = 2 + . Now 3 = 2 +  ⇒ 9 = 2 +  ⇔  = 7. 52. (a) (i) lim () = lim  = 1 →1− →1− (ii) lim () = lim (2 − 2 ) = 2 − 12 = 1. Since lim () = 1 and lim () = 1, we have lim () = 1. →1+ →1− →1+ →1+ →1 Note that the fact (1) = 3 does not affect the value of the limit. (iii) When  = 1, () = 3, so (1) = 3. (iv) lim () = lim (2 − 2 ) = 2 − 22 = 2 − 4 = −2 →2− →2− (v) lim () = lim ( − 3) = 2 − 3 = −1 →2+ →2+ (vi) lim () does not exist since lim () 6= lim (). →2 (b)       3 () = 2   2 −    −3 →2− →2+ if   1 if  = 1 if 1   ≤ 2 if   2 53. (a) (i) [[]] = −2 for −2 ≤   −1, so →−2+ (ii) [[]] = −3 for −3 ≤   −2, so →−2− lim [[]] = lim [[]] = lim (−2) = −2 →−2+ lim (−3) = −3. →−2− The right and left limits are different, so lim [[]] does not exist. →−2 (iii) [[]] = −3 for −3 ≤   −2, so lim [[]] = →−24 lim (−3) = −3. →−24 (b) (i) [[]] =  − 1 for  − 1 ≤   , so lim [[]] = lim ( − 1) =  − 1. →− →− (ii) [[]] =  for  ≤    + 1, so lim [[]] = lim  = . →+ →+ (c) lim [[]] exists ⇔  is not an integer. → 54. (a) See the graph of  = cos . Since −1 ≤ cos   0 on [− −2), we have  = () = [[cos ]] = −1 on [− −2). Since 0 ≤ cos   1 on [−2 0) ∪ (0 2], we have () = 0 on [−2 0) ∪ (0 2]. Since −1 ≤ cos   0 on (2 ], we have  () = −1 on (2 ]. Note that (0) = 1. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ (b) (i) lim  () = 0 and lim  () = 0, so lim  () = 0. →0− →0 →0+ (ii) As  → (2)− ,  () → 0, so lim →(2)− (iii) As  → (2)+ ,  () → −1, so lim  () = 0. →(2)+  () = −1. (iv) Since the answers in parts (ii) and (iii) are not equal, lim  () does not exist. →2 (c) lim  () exists for all  in the open interval (− ) except  = −2 and  = 2. → 55. The graph of  () = [[]] + [[−]] is the same as the graph of () = −1 with holes at each integer, since  () = 0 for any integer . Thus, lim  () = −1 and lim  () = −1, so lim  () = −1. However, →2− →2 →2+  (2) = [[2]] + [[−2]] = 2 + (−2) = 0, so lim  () 6=  (2). →2 56. lim →−  0  1−  2 2 √ = 0 1 − 1 = 0. As the velocity approaches the speed of light, the length approaches 0. A left-hand limit is necessary since  is not defined for   . 57. Since () is a polynomial, () = 0 + 1  + 2 2 + · · · +   . Thus, by the Limit Laws,   lim () = lim 0 + 1  + 2 2 + · · · +   = 0 + 1 lim  + 2 lim 2 + · · · +  lim  → → → 2 → →  = 0 + 1  + 2  + · · · +   = () Thus, for any polynomial , lim () = (). → 58. Let () = () where () and () are any polynomials, and suppose that () 6= 0. Then () lim () () → = → () lim  () lim () = lim → [Limit Law 5] = →  59. lim [() − 8] = lim →1 →1 () () [Exercise 57] = ().  () − 8  () − 8 · ( − 1) = lim · lim ( − 1) = 10 · 0 = 0. →1 →1 −1 −1 Thus, lim  () = lim {[ () − 8] + 8} = lim [ () − 8] + lim 8 = 0 + 8 = 8. →1 →1 Note: The value of lim →1 lim →1  () − 8 exists. −1 60. (a) lim  () = lim →0 →0  →1 →1  () − 8 does not affect the answer since it’s multiplied by 0. What’s important is that −1   () 2  () = lim 2 · lim 2 = 5 · 0 = 0 ·  →0  →0 2    ()  ()  () = lim ·  = lim 2 · lim  = 5 · 0 = 0 →0  →0 →0  →0 2 (b) lim 61. Observe that 0 ≤  () ≤ 2 for all , and lim 0 = 0 = lim 2 . So, by the Squeeze Theorem, lim  () = 0. →0 →0 →0 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 89 90 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 62. Let () = [[]] and () = −[[]]. Then lim () and lim () do not exist →3 →3 [Example 10] but lim [ () + ()] = lim ([[]] − [[]]) = lim 0 = 0. →3 →3 →3 63. Let () = () and () = 1 − (), where  is the Heaviside function defined in Exercise 1.3.59. Thus, either  or  is 0 for any value of . Then lim  () and lim () do not exist, but lim [ ()()] = lim 0 = 0. →0 →0 →0 →0 √ √ √  √ 6−−2 6−−2 6−+2 3−+1 √ √ √ √ = lim · · 64. lim →2 →2 3−−1 3−−1 6−+2 3−+1   √ 2 √ √   6 −  − 22 6−−4 3−+1 3−+1 √ √ = lim · = lim √ ·  2 →2 →2 3−−1 6−+2 6−+2 3 −  − 12 √  √ (2 − ) 3 −  + 1 1 3−+1 √  = lim √ = →2 (2 − ) →2 2 6−+2 6−+2 = lim 65. Since the denominator approaches 0 as  → −2, the limit will exist only if the numerator also approaches 0 as  → −2. In order for this to happen, we need lim →−2  2  3 +  +  + 3 = 0 ⇔ 3(−2)2 + (−2) +  + 3 = 0 ⇔ 12 − 2 +  + 3 = 0 ⇔  = 15. With  = 15, the limit becomes 3(−2 + 3) 3( + 2)( + 3) 3( + 3) 3 32 + 15 + 18 = lim = lim = = = −1. →−2 →−2 ( − 1)( + 2) →−2  − 1 2 +  − 2 −2 − 1 −3 lim 66. Solution 1: First, we find the coordinates of  and  as functions of . Then we can find the equation of the line determined by these two points, and thus find the -intercept (the point ), and take the limit as  → 0. The coordinates of  are (0 ). The point  is the point of intersection of the two circles 2 +  2 = 2 and ( − 1)2 +  2 = 1. Eliminating  from these equations, we get 2 − 2 = 1 − ( − 1)2 2 = 1 + 2 − 1 ⇔  = 12 2 . Substituting back into the equation of the ⇔ shrinking circle to find the -coordinate, we get  1 2 2  +  2 = 2 2 (the positive -value). So the coordinates of  are  − =  1 − 14 2 −  1 2  −0 2  1 2   2    ⇔  =  1 − 14 2 ⇔  2 = 2 1 − 14 2   1 − 14 2 . The equation of the line joining  and  is thus ( − 0). We set  = 0 in order to find the -intercept, and get 1 2  2  = −   =  1 − 14 2 − 1 − 12 2   1 − 14 2 + 1 1 − 14 2 − 1 =2   1 − 14 2 + 1   √  1 − 14 2 + 1 = lim 2 1 + 1 = 4. Now we take the limit as  → 0+ : lim  = lim 2 →0+ →0+ →0+ So the limiting position of  is the point (4 0). c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT ¤ 91 Solution 2: We add a few lines to the diagram, as shown. Note that ∠  = 90◦ (subtended by diameter  ). So ∠ = 90◦ = ∠ (subtended by diameter  ). It follows that ∠ = ∠ . Also ∠  = 90◦ − ∠  = ∠ . Since 4 is isosceles, so is 4 , implying that  =  . As the circle 2 shrinks, the point  plainly approaches the origin, so the point  must approach a point twice as far from the origin as  , that is, the point (4 0), as above. 2.4 The Precise Definition of a Limit 1. If |() − 1|  02, then −02   () − 1  02 ⇒ 08   ()  12. From the graph, we see that the last inequality is true if 07    11, so we can choose  = min {1 − 07 11 − 1} = min {03 01} = 01 (or any smaller positive number). 2. If |() − 2|  05, then −05   () − 2  05 ⇒ 15   ()  25. From the graph, we see that the last inequality is true if 26    38, so we can take  = min {3 − 26 38 − 3} = min {04 08} = 04 (or any smaller positive number). Note that  6= 3. 3. The leftmost question mark is the solution of √ √  = 16 and the rightmost,  = 24. So the values are 162 = 256 and 242 = 576. On the left side, we need | − 4|  |256 − 4| = 144. On the right side, we need | − 4|  |576 − 4| = 176. To satisfy both conditions, we need the more restrictive condition to hold — namely, | − 4|  144. Thus, we can choose  = 144, or any smaller positive number. 4. The leftmost question mark is the positive solution of 2 = 12 , that is,  = √12 , and the rightmost question mark is the positive solution of 2 = 32 , that is,  =      3 . On the left side, we need | − 1|   √12 − 1 ≈ 0292 (rounding down to be safe). On 2     the right side, we need | − 1|   32 − 1 ≈ 0224. The more restrictive of these two conditions must apply, so we choose  = 0224 (or any smaller positive number). 5. From the graph, we find that  = tan  = 08 when  ≈ 0675, so  −  1 ≈ 0675 4 ⇒  1 ≈ 4 − 0675 ≈ 01106. Also,  = tan  = 12 when  ≈ 0876, so 4 + 2 ≈ 0876 ⇒ 2 = 0876 − 4 ≈ 00906. Thus, we choose  = 00906 (or any smaller positive number) since this is the smaller of 1 and  2 . 6. From the graph, we find that  = 2(2 + 4) = 03 when  = 23 , so 1 −  1 = 23 ⇒  1 = 13 . Also,  = 2(2 + 4) = 05 when  = 2, so 1 +  2 = 2 ⇒  2 = 1. Thus, we choose  = 13 (or any smaller positive number) since this is the smaller of  1 and  2 . c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 92 ¤ CHAPTER 2 LIMITS AND DERIVATIVES From the graph with  = 02, we find that  = 3 − 3 + 4 = 58 when 7.  ≈ 19774, so 2 − 1 ≈ 19774 ⇒ 1 ≈ 00226. Also,  = 3 − 3 + 4 = 62 when  ≈ 2022, so 2 +  2 ≈ 20219 ⇒  2 ≈ 00219. Thus, we choose  = 00219 (or any smaller positive number) since this is the smaller of  1 and  2 . For  = 01, we get  1 ≈ 00112 and  2 ≈ 00110, so we choose  = 0011 (or any smaller positive number). From the graph with  = 05, we find that  = (2 − 1) = 15 when 8.  ≈ −0303, so  1 ≈ 0303. Also,  = (2 − 1) = 25 when  ≈ 0215, so  2 ≈ 0215. Thus, we choose  = 0215 (or any smaller positive number) since this is the smaller of  1 and  2 . For  = 01, we get  1 ≈ 0052 and  2 ≈ 0048, so we choose  = 0048 (or any smaller positive number). 9. (a) The first graph of  = 1 shows a vertical asymptote at  = 2. The second graph shows that  = 100 when ln( − 1)  ≈ 201 (more accurately, 201005). Thus, we choose  = 001 (or any smaller positive number). (b) From part (a), we see that as  gets closer to 2 from the right,  increases without bound. In symbols, lim 1 →2+ ln( − 1) = ∞. 10. We graph  = csc2  and  = 500. The graphs intersect at  ≈ 3186, so we choose  = 3186 −  ≈ 0044. Thus, if 0  | − |  0044, then csc2   500. Similarly, for  = 1000, we get  = 3173 −  ≈ 0031. 11. (a)  = 2 and  = 1000 cm2 ⇒ 2 = 1000 ⇒ 2 = 1000  ⇒ =  1000  (  0) ≈ 178412 cm. (b) | − 1000| ≤ 5 ⇒ −5 ≤ 2 − 1000 ≤ 5 ⇒ 1000 − 5 ≤ 2 ≤ 1000 + 5 ⇒       995 1005 1000 995 1005 1000 ≤  ≤ ⇒ 177966 ≤  ≤ 178858. − ≈ 004466 and − ≈ 004455. So       if the machinist gets the radius within 00445 cm of 178412, the area will be within 5 cm2 of 1000. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.4 ¤ THE PRECISE DEFINITION OF A LIMIT 93 (c)  is the radius,  () is the area,  is the target radius given in part (a),  is the target area (1000 cm2 ),  is the magnitude of the error tolerance in the area (5 cm2 ), and  is the tolerance in the radius given in part (b). 12. (a)  = 012 + 2155 + 20 and  = 200 ⇒ 012 + 2155 + 20 = 200 ⇒ [by the quadratic formula or from the graph]  ≈ 330 watts (  0) (b) From the graph, 199 ≤  ≤ 201 ⇒ 3289    3311. (c)  is the input power,  () is the temperature,  is the target input power given in part (a),  is the target temperature (200),  is the tolerance in the temperature (1), and  is the tolerance in the power input in watts indicated in part (b) (011 watts). 13. (a) |4 − 8| = 4 | − 2|  01 ⇔ | − 2|  01 01 , so  = = 0025. 4 4 (b) |4 − 8| = 4 | − 2|  001 ⇔ | − 2|  001 001 , so  = = 00025. 4 4 14. |(5 − 7) − 3| = |5 − 10| = |5( − 2)| = 5 | − 2|. We must have |() − |  , so 5 | − 2|   ⇔ | − 2|  5. Thus, choose  = 5. For  = 01,  = 002; for  = 005,  = 001; for  = 001,  = 0002. 15. Given   0, we need   0 such that if 0  | − 3|  , then       (1 + 1 ) − 2  . But (1 + 1 ) − 2   ⇔  1  − 1   ⇔ 3 3 3 1   | − 3|   ⇔ | − 3|  3. So if we choose  = 3, then 3   0  | − 3|   ⇒ (1 + 13 ) − 2  . Thus, lim (1 + 13 ) = 2 by →3 the definition of a limit. 16. Given   0, we need   0 such that if 0  | − 4|  , then |(2 − 5) − 3|  . But |(2 − 5) − 3|   ⇔ |2 − 8|   ⇔ |2| | − 4|   ⇔ | − 4|  2. So if we choose  = 2, then 0  | − 4|   ⇒ |(2 − 5) − 3|  . Thus, lim (2 − 5) = 3 by the →4 definition of a limit. 17. Given   0, we need   0 such that if 0  | − (−3)|  , then |(1 − 4) − 13|  . But |(1 − 4) − 13|   ⇔ |−4 − 12|   ⇔ |−4| | + 3|   ⇔ | − (−3)|  4. So if we choose  = 4, then 0  | − (−3)|   ⇒ |(1 − 4) − 13|  . Thus, lim (1 − 4) = 13 by the definition of a limit. →−3 x c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 94 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 18. Given   0, we need   0 such that if 0  | − (−2)|  , then |(3 + 5) − (−1)|  . But |(3 + 5) − (−1)|   ⇔ |3 + 6|   ⇔ |3| | + 2|   ⇔ | + 2|  3. So if we choose  = 3, then 0  | + 2|   ⇒ |(3 + 5) − (−1)|  . Thus, lim (3 + 5) = −1 by the definition of a limit. →−2   2 + 4 19. Given   0, we need   0 such that if 0  | − 1|  , then      2 + 4   − 2  . But  − 2   ⇔ 3 3   4  4 − 4  3 3      3    ⇔ 3 | − 1|   ⇔ | − 1|  4 . So if we choose  = 4 , then 0  | − 1|       2 + 4   . Thus, lim 2 + 4 = 2 by the definition of a limit.  − 2   3 →1 3    ⇒  20. Given   0, we need   0 such that if 0  | − 10|  , then 3 − 45  − (−5)  . But 3 − 45  − (−5)       8 − 4    ⇔ − 4  | − 10|   ⇔ | − 10|  5 . So if we choose  = 5 , then 0  | − 10|   5 5 4 4   3 − 4  − (−5)  . Thus, lim (3 − 4 ) = −5 by the definition of a limit. 5 5 ⇔ ⇒ →10  2   − 2 − 8 21. Given   0, we need   0 such that if 0  | − 4|  , then    − 6   ⇔ −4    ( − 4)( + 2)   − 6   ⇔ | + 2 − 6|   [ 6= 4] ⇔ | − 4|  . So choose  = . Then  −4    ( − 4)( + 2)  0  | − 4|   ⇒ | − 4|   ⇒ | + 2 − 6|   ⇒  − 6   [ 6= 4] ⇒ −4  2    − 2 − 8  2 − 2 − 8  − 6  . By the definition of a limit, lim = 6.  →4 −4 −4   9 − 42 22. Given   0, we need   0 such that if 0  | + 15|  , then    − 6   ⇔ 3 + 2    (3 + 2)(3 − 2)   − 6   ⇔ |3 − 2 − 6|   [ 6= −15] ⇔ |−2 − 3|   ⇔ |−2| | + 15|   ⇔  3 + 2 | + 15|  2. So choose  = 2. Then 0  | + 15|   ⇒ | + 15|  2 ⇒ |−2| | + 15|   ⇒      9 − 42    (3 + 2)(3 − 2) |−2 − 3|   ⇒ |3 − 2 − 6|   ⇒  − 6   [ 6= −15] ⇒  − 6  . 3 + 2 3 + 2 By the definition of a limit, 9 − 42 = 6. →−15 3 + 2 lim 23. Given   0, we need   0 such that if 0  | − |  , then | − |  . So  =  will work. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT ¤ 95 24. Given   0, we need   0 such that if 0  | − |  , then | − |  . But | − | = 0, so this will be true no matter what  we pick.   ⇔ 2     √ √ ⇔ ||3   ⇔ ||  3 . Take  = 3 .   25. Given   0, we need   0 such that if 0  | − 0|  , then 2 − 0   Then 0  | − 0|     ⇒ 2 − 0  . Thus, lim 2 = 0 by the definition of a limit. √ √ . Take  = . →0 26. Given   0, we need   0 such that if 0  | − 0|  , then 3 − 0   Then 0  | − 0|   ⇔ ||    ⇒ 3 − 0   3 = . Thus, lim 3 = 0 by the definition of a limit. →0   27. Given   0, we need   0 such that if 0  | − 0|  , then || − 0  . But || = ||. So this is true if we pick  = . Thus, lim || = 0 by the definition of a limit. →0 √ √   28. Given   0, we need   0 such that if 0   − (−6)  , then  8 6 +  − 0  . But  8 6 +  − 0   √ 8 6 +    ⇔ 6 +   8 ⇔  − (−6)  8 . So if we choose  = 8 , then 0   − (−6)    √ √  8 6 +  − 0  . Thus, lim 8 6 +  = 0 by the definition of a right-hand limit. ⇔ ⇒ →−6+     ⇔ 2 − 4 + 4   ⇔   √ ⇔ | − 2|   ⇔ ( − 2)2   . Thus,  29. Given   0, we need   0 such that if 0  | − 2|  , then  2 − 4 + 5 − 1     ( − 2)2   . So take  = √. Then 0  | − 2|     lim 2 − 4 + 5 = 1 by the definition of a limit. →2     30. Given   0, we need   0 such that if 0  | − 2|  , then (2 + 2 − 7) − 1  . But (2 + 2 − 7) − 1   ⇔   2  + 2 − 8   ⇔ | + 4| | − 2|  . Thus our goal is to make | − 2| small enough so that its product with | + 4| is less than . Suppose we first require that | − 2|  1. Then −1   − 2  1 ⇒ 1    3 ⇒ 5   + 4  7 ⇒ | + 4|  7, and this gives us 7 | − 2|   ⇒ | − 2|  7. Choose  = min {1 7}. Then if 0  | − 2|  , we   have | − 2|  7 and | + 4|  7, so (2 + 2 − 7) − 1 = |( + 4)( − 2)| = | + 4| | − 2|  7(7) = , as desired. Thus, lim (2 + 2 − 7) = 1 by the definition of a limit. →2    31. Given   0, we need   0 such that if 0  | − (−2)|  , then  2 − 1 − 3   or upon simplifying we need  2   − 4   whenever 0  | + 2|  . Notice that if | + 2|  1, then −1   + 2  1 ⇒ −5   − 2  −3 ⇒ | − 2|  5. So take  = min {5 1}. Then 0  | + 2|   ⇒ | − 2|  5 and | + 2|  5, so   2    − 1 − 3 = |( + 2)( − 2)| = | + 2| | − 2|  (5)(5) = . Thus, by the definition of a limit, lim (2 − 1) = 3. →−2        32. Given   0, we need   0 such that if 0  | − 2|  , then 3 − 8  . Now 3 − 8 = ( − 2) 2 + 2 + 4 . If | − 2|  1, that is, 1    3, then 2 + 2 + 4  32 + 2(3) + 4 = 19 and so   3      − 8 = | − 2| 2 + 2 + 4  19 | − 2|. So if we take  = min 1  , then 0  | − 2|   19  3     − 8 = | − 2| 2 + 2 + 4   · 19 = . Thus, by the definition of a limit, lim 3 = 8. 19 ⇒ →2 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 96 ¤ CHAPTER 2 LIMITS AND DERIVATIVES   33. Given   0, we let  = min 2 8 . If 0  | − 3|  , then | − 3|  2 ⇒ −2   − 3  2 ⇒   4   + 3  8 ⇒ | + 3|  8. Also | − 3|  8 , so 2 − 9 = | + 3| | − 3|  8 · 8 = . Thus, lim 2 = 9. →3 34. From the figure, our choices for  are  1 = 3 − 2 = √ 9 −  and √ 9 +  − 3. The largest possible choice for  is the minimum value of { 1   2 }; that is,  = min{ 1   2 } =  2 = √ 9 +  − 3. 35. (a) The points of intersection in the graph are (1  26) and (2  34) with 1 ≈ 0891 and 2 ≈ 1093. Thus, we can take  to be the smaller of 1 − 1 and 2 − 1. So  = 2 − 1 ≈ 0093. (b) Solving 3 +  + 1 = 3 +  gives us two nonreal complex roots and one real root, which is √ 23  216 + 108 + 12 336 + 324 + 812 − 12 () = √ 13 . Thus,  = () − 1.  2 6 216 + 108 + 12 336 + 324 + 81 (c) If  = 04, then () ≈ 1093 272 342 and  = () − 1 ≈ 0093, which agrees with our answer in part (a).   1 1   36. 1. Guessing a value for  Let   0 be given. We have to find a number   0 such that  −    whenever  2     1 | − 2| 1   2 −   1 0  | − 2|  . But  −  =  =  . We find a positive constant  such that  ⇒  2 2  |2| |2| | − 2|    | − 2| and we can make  | − 2|   by taking | − 2|  = . We restrict  to lie in the interval |2|  | − 2|  1 ⇒ 1    3 so 1  1 1   3 ⇒ 1 1 1   6 2 2 ⇒ 1 1 1  . So  = is suitable. Thus, we should |2| 2 2 choose  = min {1 2}. 2. Showing that  works Given   0 we let  = min {1 2}. If 0  | − 2|  , then | − 2|  1 ⇒ 1    3 ⇒   1 1 1  | − 2| 1 1  (as in part 1). Also | − 2|  2, so  −  =  · 2 = . This shows that lim (1) = 12 . →2 |2| 2  2 |2| 2 37. 1. Guessing a value for  √ √ Given   0, we must find   0 such that |  − |   whenever 0  | − |  . But √ √ √ √ | − | √   (from the hint). Now if we can find a positive constant  such that  +    then |  − | = √ +  c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT ¤ 97 | − | | − | √ √   , and we take | − |  . We can find this number by restricting  to lie in some interval  +   √ √ √ centered at . If | − |  12 , then − 12    −   12  ⇒ 12     32  ⇒  +   12  + , and so =  √ 1  is a suitable choice for the constant. So | − |  2 +  = min  1  2     . √ 1 +  2 2. Showing that  works | − |  12  ⇒ Given   0, we let  = min  1  2    . This suggests that we let √ 1  2 +    . If 0  | − |  , then √ 1 +  2   √ √ √ √  1  +   12  +  (as in part 1). Also | − |   +  , so 2  √  2 +   √ √ √ √ | − | √   |  − | = √ lim  =  by the definition of a limit. √  = . Therefore, → +  2 +  38. Suppose that lim () = . Given  = 12 , there exists   0 such that 0  ||   →0  − 12  ()   + 12 . For 0    , () = 1, so 1   + 12 ⇒ |() − |  12 ⇔ ⇒   12 . For −    0, () = 0, so  − 12  0 ⇒   12 . This contradicts   12 . Therefore, lim () does not exist. →0 39. Suppose that lim  () = . Given  = 12 , there exists   0 such that 0  ||   →0 ⇒ |() − |  12 . Take any rational number  with 0  ||  . Then  () = 0, so |0 − |  12 , so  ≤ ||  12 . Now take any irrational number  with 0  ||  . Then  () = 1, so |1 − |  12 . Hence, 1 −   12 , so   12 . This contradicts   12 , so lim () does not →0 exist. 40. First suppose that lim  () = . Then, given   0 there exists   0 so that 0  | − |   → ⇒ |() − |  . Then  −      ⇒ 0  | − |   so | () − |  . Thus, lim  () = . Also      +  →− ⇒ 0  | − |   so | () − |  . Hence, lim  () = . →+ Now suppose lim  () =  = lim (). Let   0 be given. Since lim  () = , there exists 1  0 so that →− →− →+  −  1     ⇒ | () − |  . Since lim  () = , there exists  2  0 so that      +  2 →+ | () − |  . Let  be the smaller of  1 and 2 . Then 0  | − |   ⇒  −  1     or      + 2 so | () − |  . Hence, lim  () = . So we have proved that lim  () =  ⇔ → 41. 1 1  10,000 ⇔ ( + 3)4  ( + 3)4 10,000 → 1 ⇔ | + 3|  √ 4 10,000 42. Given   0, we need   0 such that 0  | + 3|   ( + 3)4  lim 1  1 →−3 ( + 3)4 ⇒ ⇔ lim () =  = lim  (). →− →+ | − (−3)|  ⇒ 1( + 3)4  . Now 1 10 1  ( + 3)4 1 1 1 . So take  = √ . Then 0  | + 3|   = √ ⇔ | + 3|  √ 4 4 4    ⇒ ⇔ 1  , so ( + 3)4 = ∞. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 98 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 43. Given   0 we need   0 so that ln    whenever 0    ; that is,  = ln    whenever 0    . This suggests that we take  =  . If 0     , then ln   ln  = . By the definition of a limit, lim ln  = −∞. →0+ 44. (a) Let  be given. Since lim  () = ∞, there exists  1  0 such that 0  | − |   1 → lim () = , there exists  2  0 such that 0  | − |   2 → smaller of 1 and 2 . Then 0  | − |   ⇒  ()   + 1 − . Since ⇒ |() − |  1 ⇒ ()   − 1. Let  be the ⇒  () + ()  ( + 1 − ) + ( − 1) = . Thus, lim [ () + ()] = ∞. → (b) Let   0 be given. Since lim () =   0, there exists  1  0 such that 0  | − |   1 → ⇒ |() − |  2 ⇒ ()  2. Since lim  () = ∞, there exists  2  0 such that 0  | − |   2 →  ()  2. Let  = min {1   2 }. Then 0  | − |   ⇒  () ()  2  = , so lim  () () = ∞. →  2 (c) Let   0 be given. Since lim () =   0, there exists 1  0 such that 0  | − |   1 → ⇒ ⇒ |() − |  −2 ⇒ ()  2. Since lim  () = ∞, there exists  2  0 such that 0  | − |   2 ⇒  ()  2. (Note that   0 and   0 ⇒ 2  0.) Let  = min {1   2 }. Then 0  | − |   ⇒ → 2   ()  2 ⇒  () ()  · = , so lim  () () = −∞. →  2 2.5 Continuity 1. From Definition 1, lim  () =  (4). →4 2. The graph of  has no hole, jump, or vertical asymptote. 3. (a)  is discontinuous at −4 since  (−4) is not defined and at −2, 2, and 4 since the limit does not exist (the left and right limits are not the same). (b)  is continuous from the left at −2 since lim  () =  (−2).  is continuous from the right at 2 and 4 since →−2− lim () =  (2) and lim  () =  (4). It is continuous from neither side at −4 since  (−4) is undefined. →2+ →4+ 4. From the graph of , we see that  is continuous on the intervals [−3 −2), (−2 −1), (−1 0], (0 1), and (1 3]. 5. The graph of  =  () must have a discontinuity at  = 2 and must show that lim  () =  (2). →2+ 6. The graph of  =  () must have discontinuities at  = −1 and  = 4. It must show that lim  () =  (−1) and lim () = (4). →−1− →4+ c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.5 7. The graph of  =  () must have a removable CONTINUITY ¤ 99 8. The graph of  =  () must have a discontinuity discontinuity (a hole) at  = 3 and a jump discontinuity at  = −2 with at  = 5. lim  () 6=  (−2) and →−2− lim  () 6= (−2). It must also show that →−2+ lim  () =  (2) and lim  () 6=  (2). →2− →2+ 9. (a) The toll is $7 between 7:00 AM and 10:00 AM and between 4:00 PM and 7:00 PM . (b) The function  has jump discontinuities at  = 7, 10, 16, and 19. Their significance to someone who uses the road is that, because of the sudden jumps in the toll, they may want to avoid the higher rates between  = 7 and  = 10 and between  = 16 and  = 19 if feasible. 10. (a) Continuous; at the location in question, the temperature changes smoothly as time passes, without any instantaneous jumps from one temperature to another. (b) Continuous; the temperature at a specific time changes smoothly as the distance due west from New York City increases, without any instantaneous jumps. (c) Discontinuous; as the distance due west from New York City increases, the altitude above sea level may jump from one height to another without going through all of the intermediate values — at a cliff, for example. (d) Discontinuous; as the distance traveled increases, the cost of the ride jumps in small increments. (e) Discontinuous; when the lights are switched on (or off ), the current suddenly changes between 0 and some nonzero value, without passing through all of the intermediate values. This is debatable, though, depending on your definition of current. 11. lim  () = lim →−1 →−1  4  + 23 =  lim  + 2 lim 3 →−1 →−1 4 By the definition of continuity,  is continuous at  = −1. 4  = −1 + 2(−1)3 = (−3)4 = 81 = (−1). lim (2 + 5) lim 2 + 5 lim  22 + 5(2) 14 2 + 5 →2 →2 →2 = = = = = (2). 12. lim () = lim →2 →2 2 + 1 lim (2 + 1) 2 lim  + lim 1 2(2) + 1 5 →2 →2 →2 By the definition of continuity,  is continuous at  = 2. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 100 ¤ CHAPTER 2 LIMITS AND DERIVATIVES √ 13. lim () = lim 2 3 2 + 1 = 2 lim →1 →1 →1   √ 3 2 + 1 = 2 lim (3 2 + 1) = 2 3 lim  2 + lim 1 →1 →1 →1  √ = 2 3(1)2 + 1 = 2 4 = 4 = (1) By the definition of continuity,  is continuous at  = 1.   √  14. lim  () = lim 34 − 5 + 3 2 + 4 = 3 lim 4 − 5 lim  + 3 lim (2 + 4) →2 →2 →2 →2 √ = 3(2) − 5(2) + 3 22 + 4 = 48 − 10 + 2 = 40 = (2) 4 →2 By the definition of continuity,  is continuous at  = 2. 15. For   4, we have √ √ lim  () = lim ( +  − 4 ) = lim  + lim  − 4 → → →  =  + lim  − lim 4 → → √ =+ −4 → [Limit Law 1] [8, 11, and 2] [8 and 7] =  () So  is continuous at  =  for every  in (4 ∞). Also, lim  () = 4 = (4), so  is continuous from the right at 4. →4+ Thus,  is continuous on [4 ∞). 16. For   −2, we have lim ( − 1) −1 → = → 3 + 6 lim (3 + 6) lim () = lim → [Limit Law 5] → = lim  − lim 1 → → = → 3 lim  + lim 6 [2 1 and 3] → −1 3 + 6 [8 and 7] Thus,  is continuous at  =  for every  in (−∞ −2); that is,  is continuous on (−∞ −2). 17.  () = 1 is discontinuous at  = −2 because  (−2) is undefined. +2 18.  () =   1 +2  1 if  6= −2 if  = −2 Here  (−2) = 1, but lim  () = −∞ and lim  () = ∞, →−2− →−2+ so lim  () does not exist and  is discontinuous at −2. →−2 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.5 19.  () =  2 lim  () = →−1+ 101 if   −1  lim  () = ¤ if  ≤ −1 +3 →−1− CONTINUITY lim ( + 3) = −1 + 3 = 2 and →−1− lim 2 = 2−1 = 12 . Since the left-hand and the →−1+ right-hand limits of  at −1 are not equal, lim  () does not exist, and →−1  is discontinuous at −1.  2  − 2 − 1 20.  () =  1 lim  () = lim →1 if  6= 1 if  = 1 2 −  →1 2 − 1 = lim ( − 1) →1 ( + 1)( − 1) = lim  →1  + 1 = 1 , 2 but  (1) = 1, so  is discontinous at 1  cos    21.  () = 0   1 − 2 if   0 if  = 0 if   0 lim  () = 1, but  (0) = 0 6= 1, so  is discontinuous at 0. →0  2  2 − 5 − 3 −3 22.  () =  6 if  6= 3 if  = 3 22 − 5 − 3 (2 + 1)( − 3) = lim = lim (2 + 1) = 7, →3 →3 →3 −3 −3 lim  () = lim →3 but  (3) = 6, so  is discontinuous at 3. 23.  () = ( − 2)( + 1) 2 −  − 2 = =  + 1 for  6= 2. Since lim  () = 2 + 1 = 3, define  (2) = 3. Then  is →2 −2 −2 continuous at 2. 24.  () = ( − 2)(2 + 2 + 4) 2 + 2 + 4 3 − 8 4+4+4 = = for  6= 2. Since lim  () = = 3, define  (2) = 3. 2 →2  −4 ( − 2)( + 2) +2 2+2 Then  is continuous at 2. 25.  () = 26. () = 22 −  − 1 is a rational function, so it is continuous on its domain, (−∞ ∞), by Theorem 5(b). 2 + 1 2 + 1 22 −  − 1 = 2 + 1 is a rational function, so it is continuous on its domain, (2 + 1)( − 1)     −∞ − 12 ∪ − 12  1 ∪ (1 ∞), by Theorem 5(b). c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 102 ¤ CHAPTER 2 LIMITS AND DERIVATIVES √ 3 √ √  √   −2 has domain −∞ 3 2 ∪ 3 2 ∞ . Now 3 − 2 is ⇒ 3 = 2 ⇒  = 3 2, so () = 3  −2 √ 3 continuous everywhere by Theorem 5(a) and  − 2 is continuous everywhere by Theorems 5(a), 7, and 9. Thus,  is 27. 3 − 2 = 0 continuous on its domain by part 5 of Theorem 4. 28. The domain of () = sin  is (−∞ ∞) since the denominator is never 0 [cos  ≥ −1 ⇒ 2 + cos  ≥ 1]. By 2 + cos  Theorems 7 and 9, sin  and cos  are continuous on R. By part 1 of Theorem 4, 2 + cos  is continuous on R and by part 5 of Theorem 4,  is continuous on R. 29. By Theorem 5(a), the polynomial 1 + 2 is continuous on R. By Theorem 7, the inverse trigonometric function arcsin  is continuous on its domain, [−1 1]. By Theorem 9, () = arcsin(1 + 2) is continuous on its domain, which is { | −1 ≤ 1 + 2 ≤ 1} = { | −2 ≤ 2 ≤ 0} = { | −1 ≤  ≤ 0} = [−1 0].   30. By Theorem 7, the trigonometric function tan  is continuous on its domain,  |  6= 2 +  . By Theorems 5(a), 7, and 9, the composite function √ tan  4 − 2 is continuous on its domain [−2 2]. By part 5 of Theorem 4, () = √ is 4 − 2 continuous on its domain, (−2 −2) ∪ (−2 2) ∪ (2 2).   +1 1 +1 is defined when ≥ 0 ⇒  + 1 ≥ 0 and   0 or  + 1 ≤ 0 and   0 ⇒   0 31. () = 1 + =    or  ≤ −1, so  has domain (−∞ −1] ∪ (0 ∞).  is the composite of a root function and a rational function, so it is continuous at every number in its domain by Theorems 7 and 9. 2 2 32. By Theorems 7 and 9, the composite function − is continuous on R. By part 1 of Theorem 4, 1 + − is continuous on R. By Theorem 7, the inverse trigonometric function tan−1 is continuous on its domain, R. By Theorem 9, the composite   2 function () = tan−1 1 + − is continuous on its domain, R. 33. The function  = 1 is discontinuous at  = 0 because the 1 + 1 left- and right-hand limits at  = 0 are different. 34. The function  = tan2  is discontinuous at  = 2 + , where  is   any integer. The function  = ln tan2  is also discontinuous   where tan2  is 0, that is, at  = . So  = ln tan2  is discontinuous at  = 2 ,  any integer. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.5 CONTINUITY ¤ 103 √ √ √ 20 − 2 is continuous on its domain, − 20 ≤  ≤ 20, the product √ √ √  () =  20 − 2 is continuous on − 20 ≤  ≤ 20. The number 2 is in that domain, so  is continuous at 2, and √ lim  () =  (2) = 2 16 = 8. 35. Because  is continuous on R and →2 36. Because  is continuous on R, sin  is continuous on R, and  + sin  is continuous on R, the composite function  () = sin( + sin ) is continuous on R, so lim  () =  () = sin( + sin ) = sin  = 0. →  37. The function  () = ln 5 − 2 1+  is continuous throughout its domain because it is the composite of a logarithm function and a rational function. For the domain of  , we must have 5 − 2  0, so the numerator and denominator must have the 1+ √ √ same sign, that is, the domain is (−∞ − 5 ] ∪ (−1 5 ]. The number 1 is in that domain, so  is continuous at 1, and lim  () =  (1) = ln →1 5−1 = ln 2. 1+1 √ 38. The function  () = 3 2 −2−4 is continuous throughout its domain because it is the composite of an exponential function, a root function, and a polynomial. Its domain is        | 2 − 2 − 4 ≥ 0 =  | 2 − 2 + 1 ≥ 5 =  | ( − 1)2 ≥ 5   √  √ √ =   | − 1| ≥ 5 = (−∞ 1 − 5 ] ∪ [1 + 5 ∞) √ The number 4 is in that domain, so  is continuous at 4, and lim  () = (4) = 3 16−8−4 = 32 = 9. →4 39.  () =  if  ≤ 1 1 − 2 if   1 ln  By Theorem 5, since  () equals the polynomial 1 − 2 on (−∞ 1],  is continuous on (−∞ 1]. By Theorem 7, since  () equals the logarithm function ln  on (1 ∞),  is continuous on (1 ∞). At  = 1, lim  () = lim (1 − 2 ) = 1 − 12 = 0 and lim  () = lim ln  = ln 1 = 0. Thus, lim  () exists and →1− →1− →1+ →1+ →1 equals 0. Also,  (1) = 1 − 12 = 0. Thus,  is continuous at  = 1. We conclude that  is continuous on (−∞ ∞). 40.  () =  sin  if   4 cos  if  ≥ 4 By Theorem 7, the trigonometric functions are continuous. Since () = sin  on (−∞ 4) and  () = cos  on √ lim sin  = sin 4 = 1 2 since the sine (4 ∞),  is continuous on (−∞ 4) ∪ (4 ∞) lim  () = →(4)− function is continuous at 4 Similarly, at 4. Thus, lim →(4) lim →(4)+  () = lim →(4)+ →(4)− √ cos  = 1 2 by continuity of the cosine function √  () exists and equals 1 2, which agrees with the value  (4). Therefore,  is continuous at 4, so  is continuous on (−∞ ∞). c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 104 ¤ CHAPTER 2  2    41.  () =    1 LIMITS AND DERIVATIVES if   −1 if − 1 ≤   1 if  ≥ 1  is continuous on (−∞ −1), (−1 1), and (1 ∞), where it is a polynomial, a polynomial, and a rational function, respectively. Now lim 2 = 1 and lim  () = →−1− →−1− lim  () = →−1+ lim  = −1, →−1+ so  is discontinuous at −1. Since  (−1) = −1,  is continuous from the right at −1. Also, lim  () = lim  = 1 and →1− 1 lim  () = lim →1+  →1+   2   42.  () = 3 −   √  →1− = 1 =  (1), so  is continuous at 1. if  ≤ 1 if 1   ≤ 4 if   4  is continuous on (−∞ 1), (1 4), and (4 ∞), where it is an exponential, a polynomial, and a root function, respectively. Now lim  () = lim 2 = 2 and lim  () = lim (3 − ) = 2. Since  (1) = 2 we have continuity at 1. Also, →1− →1− →1+ →1+ lim  () = lim (3 − ) = −1 =  (4) and lim () = lim →4− →4− →4+ →4+ √  = 2, so  is discontinuous at 4, but it is continuous from the left at 4.  +2    43.  () =     2− if   0 if 0 ≤  ≤ 1 if   1  is continuous on (−∞ 0) and (1 ∞) since on each of these intervals it is a polynomial; it is continuous on (0 1) since it is an exponential. Now lim  () = lim ( + 2) = 2 and lim  () = lim  = 1, so  is discontinuous at 0. Since  (0) = 1,  is →0− →0− →0+ →0+ continuous from the right at 0. Also lim  () = lim  =  and lim  () = lim (2 − ) = 1, so  is discontinuous →1− →1− →1+ →1+ at 1. Since  (1) = ,  is continuous from the left at 1. 44. By Theorem 5, each piece of  is continuous on its domain. We need to check for continuity at  = . lim  () = lim →− →−       = and lim  () = lim = , so lim  () = . Since  () = , → 3 2 2 2 2 →+ →+  2  is continuous at . Therefore,  is a continuous function of . 45.  () =  2 + 2 if   2 3 −  if  ≥ 2  is continuous on (−∞ 2) and (2 ∞). Now lim () = lim →2− →2−  2   + 2 = 4 + 4 and c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.5 lim  () = lim →2+ →2+ CONTINUITY ¤ 105  3   −  = 8 − 2. So  is continuous ⇔ 4 + 4 = 8 − 2 ⇔ 6 = 4 ⇔  = 23 . Thus, for  to be continuous on (−∞ ∞),  = 23 . 46.  () =  2  −4    −2  if   2 2  −  + 3     2 −  +  At  = 2: if 2 ≤   3 if  ≥ 3 lim  () = lim →2− →2− 2 − 4 ( + 2)( − 2) = lim = lim ( + 2) = 2 + 2 = 4 −2 −2 →2− →2− lim  () = lim (2 −  + 3) = 4 − 2 + 3 →2+ →2+ We must have 4 − 2 + 3 = 4, or 4 − 2 = 1 (1). At  = 3: lim  () = lim (2 −  + 3) = 9 − 3 + 3 →3− →3− lim  () = lim (2 −  + ) = 6 −  +  →3+ →3+ We must have 9 − 3 + 3 = 6 −  + , or 10 − 4 = 3 (2). Now solve the system of equations by adding −2 times equation (1) to equation (2). −8 + 4 = −2 10 − 4 = 2 = 3 1 So  = 12 . Substituting 12 for  in (1) gives us −2 = −1, so  = 12 as well. Thus, for  to be continuous on (−∞ ∞),  =  = 12 . 47. If  and  are continuous and (2) = 6, then lim [3 () +  () ()] = 36 →2 ⇒ 3 lim  () + lim  () · lim () = 36 ⇒ 3 (2) + (2) · 6 = 36 ⇒ 9 (2) = 36 ⇒  (2) = 4. →2 →2 48. (a)  () = →2 1 1 and () = 2 , so ( ◦ )() =  (()) =  (12 ) = 1 (12 ) = 2 .   (b) The domain of  ◦  is the set of numbers  in the domain of  (all nonzero reals) such that () is in the domain of  (also     1 all nonzero reals). Thus, the domain is    6= 0 and 2 6= 0 = { |  6= 0} or (−∞ 0) ∪ (0 ∞). Since  ◦  is  the composite of two rational functions, it is continuous throughout its domain; that is, everywhere except  = 0. 49. (a)  () = (2 + 1)(2 − 1) (2 + 1)( + 1)( − 1) 4 − 1 = = = (2 + 1)( + 1) [or 3 + 2 +  + 1] −1 −1 −1 for  6= 1. The discontinuity is removable and () = 3 + 2 +  + 1 agrees with  for  6= 1 and is continuous on R. (b)  () = (2 −  − 2) ( − 2)( + 1) 3 − 2 − 2 = = = ( + 1) [or 2 + ] for  6= 2. The discontinuity −2 −2 −2 is removable and () = 2 +  agrees with  for  6= 2 and is continuous on R. (c) lim  () = lim [[sin ]] = lim 0 = 0 and lim  () = lim [[sin ]] = lim (−1) = −1, so lim  () does not →− → − → − → + → + → + → exist. The discontinuity at  =  is a jump discontinuity. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 106 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 50.  does not satisfy the conclusion of the  does satisfy the conclusion of the Intermediate Value Theorem. Intermediate Value Theorem. 51.  () = 2 + 10 sin  is continuous on the interval [31 32],  (31) ≈ 957, and (32) ≈ 1030. Since 957  1000  1030, there is a number c in (31 32) such that  () = 1000 by the Intermediate Value Theorem. Note: There is also a number c in (−32 −31) such that  () = 1000 52. Suppose that (3)  6. By the Intermediate Value Theorem applied to the continuous function  on the closed interval [2 3], the fact that  (2) = 8  6 and (3)  6 implies that there is a number  in (2 3) such that  () = 6. This contradicts the fact that the only solutions of the equation  () = 6 are  = 1 and  = 4. Hence, our supposition that  (3)  6 was incorrect. It follows that (3) ≥ 6. But  (3) 6= 6 because the only solutions of  () = 6 are  = 1 and  = 4. Therefore,  (3)  6. 53.  () = 4 +  − 3 is continuous on the interval [1 2]  (1) = −1, and (2) = 15. Since −1  0  15, there is a number  in (1 2) such that  () = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation 4 +  − 3 = 0 in the interval (1 2) √ √ √  is equivalent to the equation ln  −  +  = 0.  () = ln  −  +  is continuous on the √ √ interval [2 3],  (2) = ln 2 − 2 + 2 ≈ 0107, and  (3) = ln 3 − 3 + 3 ≈ −0169. Since  (2)  0   (3), there is a 54. The equation ln  =  − number  in (2 3) such that  () = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation √ √ ln  −  +  = 0, or ln  =  − , in the interval (2 3). 55. The equation  = 3 − 2 is equivalent to the equation  + 2 − 3 = 0.  () =  + 2 − 3 is continuous on the interval [0 1],  (0) = −2, and  (1) =  − 1 ≈ 172. Since −2  0   − 1, there is a number  in (0 1) such that  () = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation  + 2 − 3 = 0, or  = 3 − 2, in the interval (0 1). 56. The equation sin  = 2 −  is equivalent to the equation sin  − 2 +  = 0.  () = sin  − 2 +  is continuous on the interval [1 2]  (1) = sin 1 ≈ 084, and  (2) = sin 2 − 2 ≈ −109. Since sin 1  0  sin 2 − 2, there is a number  in (1 2) such that  () = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation sin  − 2 +  = 0, or sin  = 2 − , in the interval (1 2). 57. (a)  () = cos  − 3 is continuous on the interval [0 1],  (0) = 1  0, and  (1) = cos 1 − 1 ≈ −046  0. Since 1  0  −046, there is a number  in (0 1) such that  () = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation cos  − 3 = 0, or cos  = 3 , in the interval (0 1). c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.5 CONTINUITY ¤ 107 (b)  (086) ≈ 0016  0 and  (087) ≈ −0014  0, so there is a root between 086 and 087, that is, in the interval (086 087). 58. (a)  () = ln  − 3 + 2 is continuous on the interval [1 2],  (1) = −1  0, and  (2) = ln 2 + 1 ≈ 17  0. Since −1  0  17, there is a number  in (1 2) such that  () = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation ln  − 3 + 2 = 0, or ln  = 3 − 2, in the interval (1 2). (b)  (134) ≈ −003  0 and  (135) ≈ 00001  0, so there is a root between 134 and 135 that is, in the interval (134 135). 59. (a) Let  () = 100−100 − 0012  Then  (0) = 100  0 and  (100) = 100−1 − 100 ≈ −632  0. So by the Intermediate Value Theorem, there is a number  in (0 100) such that  () = 0. This implies that 100−100 = 0012 . (b) Using the intersect feature of the graphing device, we find that the root of the equation is  = 70347, correct to three decimal places. 60. (a) Let  () = arctan  +  − 1. Then  (0) = −1  0 and  (1) = 4  0. So by the Intermediate Value Theorem, there is a number  in (0 1) such that  () = 0. This implies that arctan  = 1 − . (b) Using the intersect feature of the graphing device, we find that the root of the equation is  = 0520, correct to three decimal places. 61. Let  () = sin 3 . Then  is continuous on [1 2] since  is the composite of the sine function and the cubing function, both of which are continuous on R. The zeros of the sine are at , so we note that 0  1    32   2  8  3, and that the  √ pertinent cube roots are related by 1  3 32  [call this value ]  2. [By observation, we might notice that  = 3  and √  = 3 2 are zeros of  .] Now (1) = sin 1  0,  () = sin 32  = −1  0, and  (2) = sin 8  0. Applying the Intermediate Value Theorem on [1 ] and then on [ 2], we see there are numbers  and  in (1 ) and ( 2) such that  () =  () = 0. Thus,  has at least two -intercepts in (1 2). 62. Let () = 2 − 3 + 1. Then  is continuous on (0 2] since  is a rational function whose domain is (0 ∞). By   inspection, we see that  14 = 17  0,  (1) = −1  0, and  (2) = 32  0. Appling the Intermediate Value Theorem on 16   1   1 and then on [1 2], we see there are numbers  and  in 14  1 and (1 2) such that  () =  () = 0. Thus,  has at 4 least two -intercepts in (0 2). c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 108 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 63. (⇒) If  is continuous at , then by Theorem 8 with () =  + , we have   lim  ( + ) =  lim ( + ) =  (). →0 →0 (⇐) Let   0. Since lim ( + ) =  (), there exists   0 such that 0  ||   →0 ⇒ | ( + ) −  ()|  . So if 0  | − |  , then | () −  ()| = | ( + ( − )) −  ()|  . Thus, lim  () =  () and so  is continuous at . → 64. lim sin( + ) = lim (sin  cos  + cos  sin ) = lim (sin  cos ) + lim (cos  sin ) →0 →0 =  →0 →0      lim sin  lim cos  + lim cos  lim sin  = (sin )(1) + (cos )(0) = sin  →0 →0 →0 →0 65. As in the previous exercise, we must show that lim cos( + ) = cos  to prove that the cosine function is continuous. →0 lim cos( + ) = lim (cos  cos  − sin  sin ) = lim (cos  cos ) − lim (sin  sin ) →0 →0 =  →0 →0      lim cos  lim cos  − lim sin  lim sin  = (cos )(1) − (sin )(0) = cos  →0 →0 →0 →0 66. (a) Since  is continuous at , lim () =  (). Thus, using the Constant Multiple Law of Limits, we have → lim ( )() = lim  () =  lim  () =  () = ( )(). Therefore,  is continuous at . → → → (b) Since  and  are continuous at , lim  () =  () and lim () = (). Since () 6= 0, we can use the Quotient Law → →     lim () ()  ()   → = = = of Limits: lim () = lim (). Thus, is continuous at . → → ()  lim () ()   → 67.  () =  0 if  is rational 1 if  is irrational is continuous nowhere. For, given any number  and any   0, the interval ( −   + ) contains both infinitely many rational and infinitely many irrational numbers. Since  () = 0 or 1, there are infinitely many numbers  with 0  | − |   and | () −  ()| = 1. Thus, lim () 6=  (). [In fact, lim  () does not even exist.] → 68. () =  0 if  is rational  if  is irrational → is continuous at 0. To see why, note that − || ≤ () ≤ ||, so by the Squeeze Theorem lim () = 0 = (0). But  is continuous nowhere else. For if  6= 0 and   0, the interval ( −   + ) contains both →0 infinitely many rational and infinitely many irrational numbers. Since () = 0 or , there are infinitely many numbers  with 0  | − |   and |() − ()|  || 2. Thus, lim () 6= (). → 69. If there is such a number, it satisfies the equation 3 + 1 =  ⇔ 3 −  + 1 = 0. Let the left-hand side of this equation be called  (). Now  (−2) = −5  0, and  (−1) = 1  0. Note also that  () is a polynomial, and thus continuous. So by the Intermediate Value Theorem, there is a number  between −2 and −1 such that () = 0, so that  = 3 + 1. 70.   + 3 = 0 ⇒ (3 +  − 2) + (3 + 22 − 1) = 0. Let () denote the left side of the last 3 + 22 − 1  +−2 equation. Since  is continuous on [−1 1], (−1) = −4  0, and (1) = 2  0, there exists a  in (−1 1) such that c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ¤ 109 () = 0 by the Intermediate Value Theorem. Note that the only root of either denominator that is in (−1 1) is √ √ (−1 + 5 )2 = , but () = (3 5 − 9)2 6= 0. Thus,  is not a root of either denominator, so () = 0 ⇒  =  is a root of the given equation. 71.  () = 4 sin(1) is continuous on (−∞ 0) ∪ (0 ∞) since it is the product of a polynomial and a composite of a trigonometric function and a rational function. Now since −1 ≤ sin(1) ≤ 1, we have −4 ≤ 4 sin(1) ≤ 4 . Because lim (−4 ) = 0 and lim 4 = 0, the Squeeze Theorem gives us lim (4 sin(1)) = 0, which equals (0). Thus,  is →0 →0 →0 continuous at 0 and, hence, on (−∞ ∞). 72. (a) lim  () = 0 and lim  () = 0, so lim  () = 0, which is  (0), and hence  is continuous at  =  if  = 0. For →0− →0+ →0   0, lim  () = lim  =  =  (). For   0, lim  () = lim (−) = − =  (). Thus,  is continuous at → → → →  = ; that is, continuous everywhere.     (b) Assume that  is continuous on the interval . Then for  ∈ , lim | ()| =  lim  () = | ()| by Theorem 8. (If  is → → an endpoint of , use the appropriate one-sided limit.) So | | is continuous on .  1 if  ≥ 0 (c) No, the converse is false. For example, the function () = is not continuous at  = 0, but |()| = 1 is −1 if   0 continuous on R. 73. Define () to be the monk’s distance from the monastery, as a function of time  (in hours), on the first day, and define () to be his distance from the monastery, as a function of time, on the second day. Let  be the distance from the monastery to the top of the mountain. From the given information we know that (0) = 0, (12) = , (0) =  and (12) = 0. Now consider the function  − , which is clearly continuous. We calculate that ( − )(0) = − and ( − )(12) = . So by the Intermediate Value Theorem, there must be some time 0 between 0 and 12 such that ( − )(0 ) = 0 ⇔ (0 ) = (0 ). So at time 0 after 7:00 AM, the monk will be at the same place on both days. 2.6 Limits at Infinity; Horizontal Asymptotes 1. (a) As  becomes large, the values of  () approach 5. (b) As  becomes large negative, the values of  () approach 3. 2. (a) The graph of a function can intersect a The graph of a function can intersect a horizontal asymptote. vertical asymptote in the sense that it can It can even intersect its horizontal asymptote an infinite meet but not cross it. number of times. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 110 ¤ CHAPTER 2 LIMITS AND DERIVATIVES (b) The graph of a function can have 0, 1, or 2 horizontal asymptotes. Representative examples are shown. No horizontal asymptote 3. (a) lim  () = −2 →∞ (d) lim  () = −∞ →3 4. (a) lim () = 2 →∞ (d) lim () = −∞ →2− 5. lim  () = −∞, →0 lim () = 5, One horizontal asymptote (b) lim  () = 2 (b) lim () = −1 (c) lim () = −∞ (e) lim () = ∞ (f ) Vertical:  = 0,  = 2; →−∞ 6. lim  () = ∞, →2 lim  () = 0, 9.  (0) = 3, →2− lim  () = ∞, →0+ lim  () = −∞, →−∞  is odd horizontal:  = −1,  = 2 lim  () = ∞, →−2+ lim  () = −∞, →∞ →2+ →0 →2+ lim  () = −5 →∞ →1 (e) Vertical:  = 1,  = 3; horizontal:  = −2,  = 2 →−2− 8. lim  () = 3, (c) lim  () = ∞ →−∞ →−∞ →∞ Two horizontal asymptotes lim () = 0, →−∞  (0) = 0 lim  () = 4, →0− lim () = −∞, lim  () = ∞, →2 lim  () = 0, →−∞ lim  () = ∞, →∞ lim  () = ∞, →0+ lim  () = −∞ →0− 10. lim  () = −∞, →3 lim  () = 2, →∞  (0) = 0,  is even lim  () = 2, →4+ 7. lim  () = −∞, lim () = −∞, →4− lim () = 3 →∞ c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ¤ 11. If  () = 22 , then a calculator gives (0) = 0,  (1) = 05, (2) = 1,  (3) = 1125,  (4) = 1,  (5) = 078125,  (6) = 05625,  (7) = 03828125,  (8) = 025,  (9) = 0158203125, (10) = 009765625,  (20) ≈ 000038147,    (50) ≈ 22204 × 10−12 ,  (100) ≈ 78886 × 10−27 . It appears that lim 22 = 0. →∞  12. (a) From a graph of () = (1 − 2) in a window of [0 10,000] by [0 02], we estimate that lim  () = 014 →∞ (to two decimal places.) (b) From the table, we estimate that lim  () = 01353 (to four decimal places.) →∞   () 10,000 100,000 1,000,000 0135 308 0135 333 0135 335 [Divide both the numerator and denominator by 2 22 − 7 (22 − 7)2 = lim 2 →∞ 5 +  − 3 →∞ (52 +  − 3)2 13. lim (the highest power of  that appears in the denominator)] 2 = lim (2 − 7 ) →∞ [Limit Law 5] lim (5 + 1 − 32 ) →∞ = lim 2 − lim (72 ) →∞ →∞ = →∞ lim 5 + lim (1) − lim (32 ) →∞ 2 − 7 lim (12 ) →∞ 5 + lim (1) − 3 lim (12 ) →∞ 14. lim →∞  2 − 7(0) 5 + 0 + 3(0) = 2 5  lim →∞  [Limit Laws 7 and 3] →∞ = 93 + 8 − 4 = 3 − 5 + 3 [Limit Laws 1 and 2] →∞ [Theorem 5] 93 + 8 − 4 3 − 5 + 3 [Limit Law 11] 9 + 82 − 43 →∞ 33 − 52 + 1   lim (9 + 82 − 43 )  →∞ = lim (33 − 52 + 1) = [Divide by 3 ] lim [Limit Law 5] →∞   lim 9 + lim (82 ) − lim (43 )  →∞ →∞ →∞ = lim (33 ) − lim (52 ) + lim 1 →∞ →∞   9 + 8 lim (12 ) − 4 lim (13 )  →∞ →∞ = 3 lim (13 ) − 5 lim (12 ) + 1 →∞ =  =  9 + 8(0) − 4(0) 3(0) − 5(0) + 1 [Limit Laws 1 and 2] →∞ [Limit Laws 7 and 3] →∞ [Theorem 5] 9 √ = 9=3 1 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 111 112 ¤ CHAPTER 2 LIMITS AND DERIVATIVES lim 3 − 2 lim 1 3 − 2(0) 3 3 − 2 (3 − 2) 3 − 2 →∞ →∞ = lim = lim = = = →∞ 2 + 1 →∞ (2 + 1) →∞ 2 + 1 lim 2 + lim 1 2+0 2 15. lim →∞ →∞ 1 − 2 (1 − 2 )3 13 − 1 = lim = lim →∞ 3 −  + 1 →∞ (3 −  + 1)3 →∞ 1 − 12 + 13 16. lim lim 13 − lim 1 →∞ = →∞ 17. →∞ lim 1 − lim 12 + lim 13 →∞ = →∞ 0−0 =0 1−0+0 lim 1 − 2 lim 12 0 − 2(0) −2 ( − 2)2 1 − 22 →−∞ →−∞ = lim =0 = lim = = →−∞ 2 + 1 →−∞ (2 + 1)2 →−∞ 1 + 12 lim 1 + lim 12 1+0 lim →−∞ 18. lim →−∞ →−∞ (43 + 62 − 2)3 4 + 6 − 23 4+0−0 43 + 62 − 2 = lim =2 = lim = 3 3 3 →−∞ →−∞ 2 − 4 + 5 (2 − 4 + 5) 2 − 42 + 53 2−0+0 √ √ 0+1  + 2 (  + 2 )2 132 + 1 = = −1 = lim = lim →∞ 2 − 2 →∞ (2 − 2 )2 →∞ 2 − 1 0−1 19. lim √   √  −   32 112 − 1 1 0−1 −  =− = lim = = lim →∞ 232 + 3 − 5 →∞ (232 + 3 − 5) 32 →∞ 2 + 312 − 532 2+0−0 2 20. lim (22 + 1)2 (22 + 1)2 4 [(22 + 1)2 ]2 = lim = lim →∞ ( − 1)2 (2 + ) →∞ [( − 1)2 (2 + )]4 →∞ [(2 − 2 + 1)2 ][(2 + )2 ] 21. lim (2 + 0)2 (2 + 12 )2 = =4 2 →∞ (1 − 2 + 1 )(1 + 1) (1 − 0 + 0)(1 + 0) = lim 2 2 2 1 = lim  = lim √ 4 4 2 4 →∞ →∞  +1  + 1 ( + 1)4 [since 2 = 22. lim √ →∞ √ 4 for   0] 1 1 =1 = √ = lim  →∞ 1+0 1 + 14  √ √ lim (1 + 46 )6 1 + 46 1 + 46 3 →∞ 23. lim = lim = →∞ →∞ (2 − 3 )3 2 − 3 lim (23 − 1) [since 3 = √ 6 for   0] →∞  lim 16 + 4 →∞ = lim (23 ) − lim 1 →∞ = →∞  lim (16 ) + lim 4 →∞ →∞ 0−1 √ 2 0+4 = = = −2 −1 −1  √ √ lim − (1 + 46 )6 1 + 46 1 + 46 3 →−∞ 24. lim = lim = →−∞ →−∞ (2 − 3 )3 2 − 3 lim (23 − 1) √ [since 3 = − 6 for   0] →−∞ =  lim − 16 + 4 →−∞ 2 lim (13 ) − lim 1 →−∞ →−∞ √ −2 − 0+4 = =2 = −1 −1 =  − lim (16 ) + lim 4 →−∞ →−∞ 2(0) − 1 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.6  ( + 32 )2 lim √ √  + 32  + 32  →∞ = lim = →∞ →∞ (4 − 1) 4 − 1 lim (4 − 1) 25. lim LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES [since  = ¤ 113 √ 2 for   0] →∞ =  lim 1 + 3 →∞ lim 4 − lim (1) →∞ = →∞  lim (1) + lim 3 →∞ →∞ 4−0 = √ √ 0+3 3 = 4 4  + 32 ( + 32 ) 1 + 3 = lim = lim →∞ 4 − 1 →∞ (4 − 1) →∞ 4 − 1 26. lim = ∞ since 1 + 3 → ∞ and 4 − 1 → 4 as  → ∞. √ √ √  2 92 +  − 3 92 +  + 3 92 +  − (3)2 √ √ = lim →∞ →∞ 92 +  + 3 92 +  + 3  2  9 +  − 92 1  = lim √ = lim √ · →∞ →∞ 92 +  + 3 92 +  + 3 1 √  27. lim 92 +  − 3 = lim →∞ 1 1  1 1 = = lim  = √ = lim  = 2 2 2 →∞ →∞ 3 + 3 6 9 + 3 9  +  + 3 9 + 1 + 3 28. √ 2  √ √   4 + 3 − 2 42 + 3 + 2 = lim 42 + 3 + 2 √ →−∞ →−∞ 42 + 3 − 2  2  2 4 + 3 − (2) 3 √ = lim √ = lim →−∞ →−∞ 42 + 3 − 2 42 + 3 − 2 3 3   = lim √ = lim 2 →−∞ →−∞ 4 + 3 − 2  − 4 + 3 − 2 3 3 =− = √ 4 − 4+0−2 lim 29. lim →∞ √ [since  = − 2 for   0] √ √ √  √  2 +  − 2 +  2 +  + 2 +  √ √ →∞ 2 +  + 2 +  √ √  2 +  − 2 +  = lim (2 + ) − (2 + ) [( − )] √ = lim √ = lim √ √  √ →∞ 2 +  + 2 +  →∞ 2 +  + 2 +   2 30. For   0, − − −  √ = = lim  = √ →∞ 2 1+0+ 1+0 1 +  + 1 +  √ √ √ √ 2 + 1  2 = . So as  → ∞, we have 2 + 1 → ∞, that is, lim 2 + 1 = ∞. →∞ 4 − 32 +  (4 − 32 + )3 = lim 3 →∞  −  + 2 →∞ (3 −  + 2)3 31. lim  divide by the highest power of  in the denominator   − 3 + 12 =∞ →∞ 1 − 12 + 23 = lim since the numerator increases without bound and the denominator approaches 1 as  → ∞. 32. lim (− + 2 cos 3) does not exist. lim − = 0, but lim (2 cos 3) does not exist because the values of 2 cos 3 →∞ →∞ →∞ oscillate between the values of −2 and 2 infinitely often, so the given limit does not exist. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 114 ¤ 33. lim (2 + 27 ) = lim 7 CHAPTER 2 →−∞ LIMITS AND DERIVATIVES →−∞   1 + 2 [factor out the largest power of ] = −∞ because 7 → −∞ and 5 15 + 2 → 2 as  → −∞.     Or: lim 2 + 27 = lim 2 1 + 25 = −∞. →−∞ 34. →−∞ 1 + 6 (1 + 6 )4 = lim →−∞ 4 + 1 →−∞ (4 + 1)4 lim  divide by the highest power of  in the denominator  = lim →−∞ 14 + 2 =∞ 1 + 14 since the numerator increases without bound and the denominator approaches 1 as  → −∞. 35. Let  =  . As  → ∞,  → ∞. lim arctan( ) = lim arctan  = 2 by (3). →∞ →∞ 1−0 3 − −3 1 − −6 =1 = lim = 3 −3 →∞  →∞ 1 + −6 + 1+0 36. Divide numerator and denominator by 3 : lim 0−1 1 1 −  (1 −  ) 1 − 1 = =− = lim = lim    →∞ 1 + 2 →∞ (1 + 2 ) →∞ 1 + 2 0+2 2 37. lim 38. Since 0 ≤ sin2  ≤ 1, we have 0 ≤ Theorem, lim sin2  →∞ 2 + 1 1 sin2  1 ≤ 2 . We know that lim 0 = 0 and lim 2 = 0, so by the Squeeze →∞ →∞  + 1 2 + 1  +1 = 0. 39. Since −1 ≤ cos  ≤ 1 and −2  0, we have −−2 ≤ −2 cos  ≤ −2 . We know that lim (−−2 ) = 0 and →∞   lim −2 = 0, so by the Squeeze Theorem, lim (−2 cos ) = 0. →∞ →∞ 40. Let  = ln . As  → 0+ ,  → −∞. lim tan−1 (ln ) = lim tan−1  = − 2 by (4). →−∞ →0+ 41. lim [ln(1 + 2 ) − ln(1 + )] = lim ln →∞ →∞ parentheses is ∞. 42. lim [ln(2 + ) − ln(1 + )] = lim ln →∞ →∞ 1 + 2 = ln 1+  2+ 1+   1 + 2 →∞ 1 +  lim = lim ln →∞   2 + 1 1 + 1 = ln   (ii) lim  () = lim  (iii) lim  () = lim  = ∞ since  → 1 and ln  → 0+ as  → 1+ . ln  →1− →1+ →0+ ln  →1− ln  →1+ lim = ln 43. (a) (i) lim  () = lim →0+  1 +  = ∞, since the limit in →∞ 1 + 1   1 = ln 1 = 0 1 = 0 since  → 0+ and ln  → −∞ as  → 0+ . = −∞ since  → 1 and ln  → 0− as  → 1− . (c) (b)   () 10,000 10857 100,000 86859 1,000,000 72,3824 It appears that lim () = ∞. →∞ c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.6 44. (a) lim  () = lim →∞ →∞  1 2 −  ln   LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ¤ (e) =0 2 1 → 0 and → 0 as  → ∞.  ln    1 2 − (b) lim () = lim =∞  ln  →0+ →0+ since 1 2 → ∞ and → 0 as  → 0+ .  ln    1 2 1 2 (c) lim  () = lim − = ∞ since → 2 and → −∞ as  → 1− .  ln   ln  →1− →1− since (d) lim () = lim →1+ →1+  1 2 −  ln   = −∞ since 1 2 → 2 and → ∞ as  → 1+ .  ln  (b) 45. (a)   () −10,000 −0499 962 5 −1,000,000 −0499 999 6 −100,000 From the graph of  () = √ 2 +  + 1 + , we −0499 996 2 From the table, we estimate the limit to be −05. estimate the value of lim  () to be −05. →−∞  2  √ 2  √ √    +  + 1 − 2  ++1− 2 +  + 1 +  = lim 2 +  + 1 +  √ = lim √ →−∞ →−∞ →−∞ 2 +  + 1 −  2 +  + 1 −  (c) lim ( + 1)(1) 1 + (1)   = lim = lim √ →−∞ 2 +  + 1 −  (1) →−∞ − 1 + (1) + (12 ) − 1 = Note that for   0, we have 1 1+0 √ =− 2 − 1+0+0−1 √ 2 = || = −, so when we divide the radical by , with   0, we get  1 √ 2 1√ 2  +  + 1 = −√  +  + 1 = − 1 + (1) + (12 ). 2   46. (a) (b) From the graph of √ √  () = 32 + 8 + 6 − 32 + 3 + 1, we estimate   () 10,000 1443 39 100,000 1443 38 1,000,000 1443 38 From the table, we estimate (to four decimal places) the limit to be 14434. (to one decimal place) the value of lim  () to be 14. →∞ c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 115 116 ¤ CHAPTER 2 LIMITS AND DERIVATIVES √ √ √ √  32 + 8 + 6 − 32 + 3 + 1 32 + 8 + 6 + 32 + 3 + 1 √ √ →∞ 32 + 8 + 6 + 32 + 3 + 1  2   2  3 + 8 + 6 − 3 + 3 + 1 (5 + 5)(1) √ √  = lim √ = lim √ →∞ →∞ 32 + 8 + 6 + 32 + 3 + 1 32 + 8 + 6 + 32 + 3 + 1 (1) √ 5 5 5 3 5 + 5  √ = √ = = lim  ≈ 1443376 = √ →∞ 6 3+ 3 2 3 3 + 8 + 62 + 3 + 3 + 12 (c) lim  () = lim →∞ 47. lim →±∞ 0+4 5 + 4 (5 + 4) 5 + 4 = lim = lim = = 4, so →±∞ ( + 3) →±∞ 1 + 3 +3 1+0  = 4 is a horizontal asymptote.  =  () = 5 + 4 , so lim  () = −∞ +3 →−3+ since 5 + 4 → −7 and  + 3 → 0+ as  → −3+ . Thus,  = −3 is a vertical asymptote. The graph confirms our work. 48. lim 22 + 1 →±∞ 32 + 2 − 1 (22 + 1)2 →±∞ (32 + 2 − 1)2 = lim 2 + 12 2 = →±∞ 3 + 2 − 12 3 = lim so  = 22 + 1 22 + 1 2 is a horizontal asymptote.  =  () = 2 = . 3 3 + 2 − 1 (3 − 1)( + 1) The denominator is zero when  = 13 and −1, but the numerator is nonzero, so  = 13 and  = −1 are vertical asymptotes. The graph confirms our work.   1 1 1 1 22 +  − 1 2 + − lim 2+ − 2  2 22 +  − 1 2   = →±∞   = lim 49. lim = lim 2 2 2 1 →±∞  +  − 2 →±∞  +  − 2 →±∞ 1 2 1+ − 2 1+ − 2 lim   →±∞   2 1 1 lim 2 + lim − lim 2+0−0 →±∞ →±∞  →±∞ 2 = 2, so  = 2 is a horizontal asymptote. = = 1 1 1 + 0 − 2(0) − 2 lim lim 1 + lim →±∞ →±∞  →±∞ 2  =  () = (2 − 1)( + 1) 22 +  − 1 = , so lim () = ∞, 2 +  − 2 ( + 2)( − 1) →−2− lim  () = −∞, lim  () = −∞, and lim  () = ∞. Thus,  = −2 →1− →−2+ →1+ and  = 1 are vertical asymptotes. The graph confirms our work.   1 1 1 + 4 1 + 1 lim lim + lim 1 +1 4 4 →±∞ 4 4  1 + 4 →±∞ →±∞    =  50. lim = lim = lim  = 2 4 2 4 1 →±∞  −  →±∞  −  →±∞ 1 1 −1 lim − lim 1 lim −1 →±∞ 2 →±∞ 2 →±∞ 2 4 = 0+1 = −1, so  = −1 is a horizontal asymptote. 0−1 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.6  =  () = LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES 1 + 4 1 + 4 1 + 4 = . The denominator is = 2 − 4 2 (1 − 2 ) 2 (1 + )(1 − ) zero when  = 0, −1, and 1, but the numerator is nonzero, so  = 0,  = −1, and  = 1 are vertical asymptotes. Notice that as  → 0, the numerator and denominator are both positive, so lim  () = ∞. The graph confirms our work. →0 51.  =  () = (2 − 1) ( + 1)( − 1) ( + 1) 3 −  = = = = () for  6= 1. 2 − 6 + 5 ( − 1)( − 5) ( − 1)( − 5) −5 The graph of  is the same as the graph of  with the exception of a hole in the graph of  at  = 1. By long division, () = 2 +  30 =+6+ . −5 −5 As  → ±∞, () → ±∞, so there is no horizontal asymptote. The denominator of  is zero when  = 5. lim () = −∞ and lim () = ∞, so  = 5 is a →5− →5+ vertical asymptote. The graph confirms our work. 52. lim 2 →∞  − 5 = lim 2 →∞  − 5 2 · 1 2 2 = = 2, so  = 2 is a horizontal asymptote. = lim →∞ 1 − (5 ) 1 1−0 2(0) = 0, so  = 0 is a horizontal asymptote. The denominator is zero (and the numerator isn’t) 0−5 when  − 5 = 0 ⇒  = 5 ⇒  = ln 5. lim →−∞  − 5 lim →(ln 5)+ = 2 = ∞ since the numerator approaches 10 and the denominator  − 5 approaches 0 through positive values as  → (ln 5)+ . Similarly, 2 = −∞. Thus,  = ln 5 is a vertical asymptote. The graph  →(ln 5)−  − 5 lim confirms our work. 53. From the graph, it appears  = 1 is a horizontal asymptote. 33 + 5002 33 + 5002 3 = lim lim →±∞ 3 + 5002 + 100 + 2000 →±∞ 3 + 5002 + 100 + 2000 3 = lim 3 + (500) →±∞ 1 + (500) + (1002 ) + (20003 ) = 3+0 = 3, so  = 3 is a horizontal asymptote. 1+0+0+0 The discrepancy can be explained by the choice of the viewing window. Try [−100,000 100,000] by [−1 4] to get a graph that lends credibility to our calculation that  = 3 is a horizontal asymptote. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° ¤ 117 118 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 54. (a) From the graph, it appears at first that there is only one horizontal asymptote, at  ≈ 0 and a vertical asymptote at  ≈ 17. However, if we graph the function with a wider and shorter viewing rectangle, we see that in fact there seem to be two horizontal asymptotes: one at  ≈ 05 and one at  ≈ −05. So we estimate that √ 22 + 1 ≈ 05 →∞ 3 − 5 √ 22 + 1 ≈ −05 →−∞ 3 − 5 √ 22 + 1 ≈ 047. (b)  (1000) ≈ 04722 and (10,000) ≈ 04715, so we estimate that lim →∞ 3 − 5 √ 22 + 1  (−1000) ≈ −04706 and (−10,000) ≈ −04713, so we estimate that lim ≈ −047. →−∞ 3 − 5 lim and lim  √ √ √ 2 + 12 22 + 1 2 2 (c) lim = lim [since  =  for   0] = ≈ 0471404. →∞ 3 − 5 →∞ 3 − 5 3 √ For   0, we have 2 = || = −, so when we divide the numerator by , with   0, we  1 √ 2 1√ 2 2 + 1 = − √ 2 + 1 = − 2 + 12 . Therefore, 2    √ √ − 2 + 12 22 + 1 2 lim = lim =− ≈ −0471404. →−∞ 3 − 5 →−∞ 3 − 5 3 get 55. Divide the numerator and the denominator by the highest power of  in (). (a) If deg   deg , then the numerator → 0 but the denominator doesn’t. So lim [ ()()] = 0. →∞ (b) If deg   deg , then the numerator → ±∞ but the denominator doesn’t, so lim [ ()()] = ±∞ →∞ (depending on the ratio of the leading coefficients of  and ). 56. (i)  = 0 (ii)   0 ( odd) (iii)   0 ( even) From these sketches we see that  1 if  = 0    (a) lim  = 0 if   0  →0+  ∞ if   0 (b) lim  = →0− (iv)   0 ( odd)       (v)   0 ( even) 1 if  = 0 0 if   0  −∞ if   0,  odd     ∞ if   0,  even c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.6  1 if  = 0   (c) lim  = ∞ if   0 →∞   0 if   0 (d) lim  = →−∞ 57. Let’s look for a rational function. (1) LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ¤ 119  1 if  = 0     −∞ if   0,  odd  ∞ if   0,  even     0 if   0 lim () = 0 ⇒ degree of numerator  degree of denominator →±∞ (2) lim  () = −∞ ⇒ there is a factor of 2 in the denominator (not just , since that would produce a sign →0 change at  = 0), and the function is negative near  = 0. (3) lim  () = ∞ and lim () = −∞ ⇒ vertical asymptote at  = 3; there is a factor of ( − 3) in the →3− →3+ denominator. (4)  (2) = 0 ⇒ 2 is an -intercept; there is at least one factor of ( − 2) in the numerator. Combining all of this information and putting in a negative sign to give us the desired left- and right-hand limits gives us  () = 2− as one possibility. 2 ( − 3) 58. Since the function has vertical asymptotes  = 1 and  = 3, the denominator of the rational function we are looking for must have factors ( − 1) and ( − 3). Because the horizontal asymptote is  = 1, the degree of the numerator must equal the degree of the denominator, and the ratio of the leading coefficients must be 1. One possibility is  () = 2 . ( − 1)( − 3) 59. (a) We must first find the function  . Since  has a vertical asymptote  = 4 and -intercept  = 1,  − 4 is a factor of the denominator and  − 1 is a factor of the numerator. There is a removable discontinuity at  = −1, so  − (−1) =  + 1 is a factor of both the numerator and denominator. Thus,  now looks like this:  () = be determined. Then lim  () = lim →−1  = 5. Thus  () =  (0) = (−1 − 1) 2 2 ( − 1)( + 1) ( − 1) = lim = = , so  = 2, and →−1  − 4 ( − 4)( + 1) (−1 − 4) 5 5 5( − 1)( + 1) is a ratio of quadratic functions satisfying all the given conditions and ( − 4)( + 1) 5(−1)(1) 5 = . (−4)(1) 4 (b) lim  () = 5 lim →∞ →−1 ( − 1)( + 1) , where  is still to ( − 4)( + 1) 2 − 1 →∞ 2 − 3 − 4 60.  =  () = 23 − 4 = 3 (2 − ). = 5 lim (2 2 ) − (12 ) →∞ (2 2 ) − (32 ) − (42 ) =5 1−0 = 5(1) = 5 1−0−0 The -intercept is (0) = 0. The -intercepts are 0 and 2. There are sign changes at 0 and 2 (odd exponents on  and 2 − ). As  → ∞,  () → −∞ because 3 → ∞ and 2 −  → −∞. As  → −∞,  () → −∞ because 3 → −∞ and 2 −  → ∞. Note that the graph of  near  = 0 flattens out (looks like  = 3 ). c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 120 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 61.  =  () = 4 − 6 = 4 (1 − 2 ) = 4 (1 + )(1 − ). The -intercept is  (0) = 0. The -intercepts are 0, −1, and 1 [found by solving  () = 0 for ]. Since 4  0 for  6= 0,  doesn’t change sign at  = 0. The function does change sign at  = −1 and  = 1. As  → ±∞,  () = 4 (1 − 2 ) approaches −∞ because 4 → ∞ and (1 − 2 ) → −∞. 62.  =  () = 3 ( + 2)2 ( − 1). The -intercept is  (0) = 0. The -intercepts are 0, −2, and 1. There are sign changes at 0 and 1 (odd exponents on  and  − 1). There is no sign change at −2. Also,  () → ∞ as  → ∞ because all three factors are large. And  () → ∞ as  → −∞ because 3 → −∞, ( + 2)2 → ∞, and ( − 1) → −∞. Note that the graph of  at  = 0 flattens out (looks like  = −3 ). 63.  =  () = (3 − )(1 + )2 (1 − )4 . The -intercept is  (0) = 3(1)2 (1)4 = 3. The -intercepts are 3, −1, and 1. There is a sign change at 3, but not at −1 and 1. When  is large positive, 3 −  is negative and the other factors are positive, so lim  () = −∞. When  is large negative, 3 −  is positive, so →∞ lim () = ∞. →−∞ 64.  =  () = 2 (2 − 1)2 ( + 2) = 2 ( + 1)2 ( − 1)2 ( + 2). The -intercept is  (0) = 0. The -intercepts are 0, −1, 1 and −2. There is a sign change at −2, but not at 0, −1, and 1. When  is large positive, all the factors are positive, so lim  () = ∞. When  is large negative, only  + 2 is negative, so →∞ lim () = −∞. →−∞ sin  1 1 ≤ ≤ for   0. As  → ∞, −1 → 0 and 1 → 0, so by the Squeeze    sin  Theorem, (sin ) → 0. Thus, lim = 0. →∞  65. (a) Since −1 ≤ sin  ≤ 1 for all  − (b) From part (a), the horizontal asymptote is  = 0. The function  = (sin ) crosses the horizontal asymptote whenever sin  = 0; that is, at  =  for every integer . Thus, the graph crosses the asymptote an infinite number of times. 66. (a) In both viewing rectangles, lim  () = lim () = ∞ and →∞ →∞ lim  () = lim () = −∞. →−∞ →−∞ In the larger viewing rectangle,  and  become less distinguishable. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ¤ 121   5 1 2 1  () 35 − 53 + 2 1 − = 1 − 53 (0) + 23 (0) = 1 ⇒ = lim · · (b) lim = lim + →∞ () →∞ →∞ 35 3 2 3 4  and  have the same end behavior. √ √ 1  5 5  5 · √ = lim  = 5 and = √ →∞  − 1 1  1−0 1 − (1) 67. lim √ →∞ √ 10 − 0 10 − 21 5  10 − 21 1 10 − (21 ) √ lim , = = 5. Since · = lim   ()  →∞ →∞ 2 1 2 2 2 −1 we have lim () = 5 by the Squeeze Theorem. →∞ 68. (a) After  minutes, 25 liters of brine with 30 g of salt per liter has been pumped into the tank, so it contains (5000 + 25) liters of water and 25 · 30 = 750 grams of salt. Therefore, the salt concentration at time  will be () = 750 30 g = . 5000 + 25 200 +  L (b) lim () = lim 30 →∞ 200 +  →∞ = lim 30 →∞ 200 +  = 30 = 30. So the salt concentration approaches that of the brine 0+1 being pumped into the tank.   ∗ =  ∗ (1 − 0) =  ∗ 69. (a) lim () = lim  ∗ 1 − − →∞ →∞ (b) We graph () = 1 − −98 and () = 099 ∗ , or in this case, () = 099. Using an intersect feature or zooming in on the point of intersection, we find that  ≈ 047 s. 70. (a)  = −10 and  = 01 intersect at 1 ≈ 2303. If   1 , then −10  01. (b) −10  01 ⇒ −10  ln 01 ⇒ 1 = −10 ln 10−1 = 10 ln 10 ≈ 2303   −10 ln 10 71. Let () = 32 + 1 and  () = |() − 15|. Note that 22 +  + 1 lim () = 32 and lim () = 0. We are interested in finding the →∞ →∞ -value at which  ()  005. From the graph, we find that  ≈ 14804, so we choose  = 15 (or any larger number). 72. We want to find a value of  such that        1 − 3 − (−3)  , or equivalently, ⇒  √ 2  +1 1 − 3 1 − 3 −3 −   √  −3 + . When  = 01, we graph  =  () = √ ,  = −31, and  = −29. From the graph, 2  +1 2 + 1 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 122 ¤ CHAPTER 2 LIMITS AND DERIVATIVES we find that () = −29 at about  = 11283, so we choose  = 12 (or any larger number). Similarly for  = 005, we find that () = −295 at about  = 21379, so we choose  = 22 (or any larger number).   1 − 3   1 − 3 − 3  , or equivalently, 3 −   √  3 + . When  = 01, 2 + 1 2 + 1 73. We want a value of  such that    ⇒  √ 1 − 3 we graph  =  () = √ ,  = 31, and  = 29. From the graph, we find that  () = 31 at about  = −8092, so we 2 + 1 choose  = −9 (or any lesser number). Similarly for  = 005, we find that  () = 305 at about  = −18338, so we choose  = −19 (or any lesser number). 74. We want to find a value of  such that    We graph  =  () = ⇒ √  ln   100. √  ln  and  = 100. From the graph, we find that () = 100 at about  = 1382773, so we choose  = 1383 (or any larger number). 75. (a) 12  00001 ⇔ 2  100001 = 10 000 ⇔   100 (  0) √ √ (b) If   0 is given, then 12   ⇔ 2  1 ⇔   1 . Let  = 1 .   1  1 1 1 ⇒  2 − 0 = 2  , so lim 2 = 0. Then    ⇒   √ →∞     √ √ 76. (a) 1   00001 ⇔   100001 = 104 ⇔   108 √ √ (b) If   0 is given, then 1    ⇔   1 ⇔   12 . Let  = 12 .     1 1 1 1 Then    ⇒   2 ⇒  √ − 0 = √  , so lim √ = 0. →∞     77. For   0, |1 − 0| = −1. If   0 is given, then −1   Take  = −1. Then    ⇒   −1 ⇒ |(1) − 0| = −1  , so lim (1) = 0. →−∞ 78. Given   0, we need   0 such that    √  = 3 ⇔   −1. ⇒ 3  . Now 3   √ √ ⇔   3 , so take  = 3 . Then ⇒ 3  , so lim 3 = ∞. →∞ c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.7 79. Given   0, we need   0 such that    DERIVATIVES AND RATES OF CHANGE ⇒   . Now    ¤ 123 ⇔   ln , so take  = max(1 ln ). (This ensures that   0.) Then    = max(1 ln ) ⇒   max( ) ≥ , so lim  = ∞. →∞ 80. Definition Let  be a function defined on some interval (−∞ ). Then lim () = −∞ means that for every negative →−∞ number  there is a corresponding negative number  such that  ()   whenever   . Now we use the definition to   prove that lim 1 + 3 = −∞. Given a negative number , we need a negative number  such that    ⇒ →−∞ √ √ ⇔ 3   − 1 ⇔   3  − 1. Thus, we take  = 3  − 1 and find that   ⇒ 1 + 3  . This proves that lim 1 + 3 = −∞. 1 + 3  . Now 1 + 3    →−∞ 81. (a) Suppose that lim  () = . Then for every   0 there is a corresponding positive number  such that | () − |   →∞ whenever   . If  = 1, then    ⇔ 0  1  1 ⇔ 0    1. Thus, for every   0 there is a corresponding   0 (namely 1) such that | (1) − |   whenever 0    . This proves that lim (1) =  = lim (). →∞ →0+ Now suppose that lim () = . Then for every   0 there is a corresponding negative number  such that →−∞ | () − |   whenever   . If  = 1, then    ⇔ 1  1  0 ⇔ 1    0. Thus, for every   0 there is a corresponding   0 (namely −1) such that | (1) − |   whenever −    0. This proves that lim  (1) =  = lim  (). →−∞ →0− (b) lim  sin →0+ 1 1 = lim  sin  →0+  = lim 1 →∞  = lim →∞ =0 sin  sin   [let  = ] [part (a) with  = 1] [let  = ] [by Exercise 65] 2.7 Derivatives and Rates of Change 1. (a) This is just the slope of the line through two points:   = ∆  () −  (3) = . ∆ −3 (b) This is the limit of the slope of the secant line   as  approaches  :  = lim →3 () −  (3) . −3 2. The curve looks more like a line as the viewing rectangle gets smaller. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 124 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 3. (a) (i) Using Definition 1 with  () = 4 − 2 and  (1 3),  = lim →  () −  () (4 − 2 ) − 3 −(2 − 4 + 3) −( − 1)( − 3) = lim = lim = lim →1 →1 →1 − −1 −1 −1 = lim (3 − ) = 3 − 1 = 2 →1 (ii) Using Equation 2 with  () = 4 − 2 and  (1 3),   4(1 + ) − (1 + )2 − 3  ( + ) − ()  (1 + ) −  (1) = lim = lim  = lim →0 →0 →0    4 + 4 − 1 − 2 − 2 − 3 −2 + 2 (− + 2) = lim = lim = lim (− + 2) = 2 →0 →0 →0 →0    = lim (b) An equation of the tangent line is  − () =  0 ()( − ) ⇒  −  (1) =  0 (1)( − 1) ⇒  − 3 = 2( − 1), or  = 2 + 1. The graph of  = 2 + 1 is tangent to the graph of  = 4 − 2 at the (c) point (1 3). Now zoom in toward the point (1 3) until the parabola and the tangent line are indistiguishable. 4. (a) (i) Using Definition 1 with  () =  − 3 and  (1 0),  () − 0  − 3 (1 − 2 ) (1 + )(1 − ) = lim = lim = lim →1  − 1 →1  − 1 →1 →1 −1 −1  = lim = lim [−(1 + )] = −1(2) = −2 →1 (ii) Using Equation 2 with  () =  − 3 and  (1 0),   (1 + ) − (1 + )3 − 0  ( + ) −  () (1 + ) − (1) = lim = lim →0 →0 →0     = lim 1 +  − (1 + 3 + 32 + 3 ) −3 − 32 − 2 (−2 − 3 − 2) = lim = lim →0 →0 →0    = lim = lim (−2 − 3 − 2) = −2 →0 (b) An equation of the tangent line is  −  () =  0 ()( − ) ⇒  −  (1) =  0 (1)( − 1) ⇒  − 0 = −2( − 1), or  = −2 + 2. (c) The graph of  = −2 + 2 is tangent to the graph of  =  − 3 at the point (1 0). Now zoom in toward the point (1 0) until the cubic and the tangent line are indistinguishable. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.7 DERIVATIVES AND RATES OF CHANGE 5. Using (1) with  () = 4 − 32 and  (2 −4) [we could also use (2)],   4 − 32 − (−4)  () −  () −32 + 4 + 4  = lim = lim = lim → →2 →2 − −2 −2 = lim →2 (−3 − 2)( − 2) = lim (−3 − 2) = −3(2) − 2 = −8 →2 −2 Tangent line:  − (−4) = −8( − 2) ⇔  + 4 = −8 + 16 ⇔  = −8 + 12. 6. Using (2) with  () = 3 − 3 + 1 and  (2 3),  = lim  ( + ) −  ()  (2 + ) −  (2) (2 + )3 − 3(2 + ) + 1 − 3 = lim = lim →0 →0    = lim (9 + 6 + 2 ) 8 + 12 + 62 + 3 − 6 − 3 − 2 9 + 62 + 3 = lim = lim →0 →0    →0 →0 = lim (9 + 6 + 2 ) = 9 →0 Tangent line:  − 3 = 9( − 2) ⇔  − 3 = 9 − 18 ⇔  = 9 − 15 √ √ √ √ 1 − 1 (  − 1)(  + 1) −1 1 √ √ = lim = lim √ = . = lim →1 →1 ( − 1)(  + 1) →1 ( − 1)(  + 1) →1 −1 2 +1 7. Using (1),  = lim Tangent line:  − 1 = 12 ( − 1) ⇔ 8. Using (1) with  () =  = 12  + 12 2 + 1 and  (1 1), +2 2 + 1 2 + 1 − ( + 2) −1  () −  () −1  + 2 +2  = lim = lim = lim = lim → →1 →1 →1 ( − 1)( + 2) − −1 −1 = lim 1 →1  + 2 = 1 1 = 1+2 3 Tangent line:  − 1 = 13 ( − 1) ⇔  − 1 = 13  − 13 ⇔  = 13  + 23 9. (a) Using (2) with  =  () = 3 + 42 − 23 ,  ( + ) −  () 3 + 4( + )2 − 2( + )3 − (3 + 42 − 23 ) = lim →0 →0    = lim 3 + 4(2 + 2 + 2 ) − 2(3 + 32  + 32 + 3 ) − 3 − 42 + 23 →0  = lim 3 + 42 + 8 + 42 − 23 − 62  − 62 − 23 − 3 − 42 + 23 →0  = lim = lim →0 8 + 42 − 62  − 62 − 23 (8 + 4 − 62 − 6 − 22 ) = lim →0   = lim (8 + 4 − 62 − 6 − 22 ) = 8 − 62 →0 (b) At (1 5):  = 8(1) − 6(1)2 = 2, so an equation of the tangent line (c) is  − 5 = 2( − 1) ⇔  = 2 + 3. At (2 3):  = 8(2) − 6(2)2 = −8, so an equation of the tangent line is  − 3 = −8( − 2) ⇔  = −8 + 19. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° ¤ 125 126 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 10. (a) Using (1), 1 1 √ −√   = lim  = lim → → − √ √ −  √ √ √ √ √ (  − )(  + ) −  √ √ √ √ = lim √ = lim √ → −  ( − ) (  +  ) →  ( − ) (  +  ) 1 −1 1 −1 √ √ = − 32 or − −32 [  0] = √ = lim √ √ → 2 2  (  +  ) 2 (2  ) (b) At (1 1):  = − 12 , so an equation of the tangent line is  − 1 = − 12 ( − 1) ⇔ (c)  = − 12  + 32 .   1 At 4 12 :  = − 16 , so an equation of the tangent line 1 1 is  − 12 = − 16 ( − 4) ⇔  = − 16  + 34 . 11. (a) The particle is moving to the right when  is increasing; that is, on the intervals (0 1) and (4 6). The particle is moving to the left when  is decreasing; that is, on the interval (2 3). The particle is standing still when  is constant; that is, on the intervals (1 2) and (3 4). (b) The velocity of the particle is equal to the slope of the tangent line of the graph. Note that there is no slope at the corner points on the graph. On the interval (0 1) the slope is 3−0 = 3. On the interval (2 3), the slope is 1−0 3−1 1−3 = −2. On the interval (4 6), the slope is = 1. 3−2 6−4 12. (a) Runner A runs the entire 100-meter race at the same velocity since the slope of the position function is constant. Runner B starts the race at a slower velocity than runner A, but finishes the race at a faster velocity. (b) The distance between the runners is the greatest at the time when the largest vertical line segment fits between the two graphs—this appears to be somewhere between 9 and 10 seconds. (c) The runners had the same velocity when the slopes of their respective position functions are equal—this also appears to be at about 95 s. Note that the answers for parts (b) and (c) must be the same for these graphs because as soon as the velocity for runner B overtakes the velocity for runner A, the distance between the runners starts to decrease. 13. Let () = 40 − 162 .     40 − 162 − 16 −8 22 − 5 + 2 () − (2) −162 + 40 − 16 = lim = lim = lim (2) = lim →2 →2 →2 →2 −2 −2 −2 −2 = lim →2 −8( − 2)(2 − 1) = −8 lim (2 − 1) = −8(3) = −24 →2 −2 Thus, the instantaneous velocity when  = 2 is −24 fts. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.7 DERIVATIVES AND RATES OF CHANGE ¤ 127 14. (a) Let () = 10 − 1862 .   10(1 + ) − 186(1 + )2 − (10 − 186) (1 + ) − (1) = lim →0 →0   (1) = lim 10 + 10 − 186(1 + 2 + 2 ) − 10 + 186 →0  = lim 10 + 10 − 186 − 372 − 1862 − 10 + 186 →0  = lim 628 − 1862 = lim (628 − 186) = 628 →0 →0  = lim The velocity of the rock after one second is 628 ms.   10( + ) − 186( + )2 − (10 − 1862 ) ( + ) − () (b) () = lim = lim →0 →0   = lim 10 + 10 − 186(2 + 2 + 2 ) − 10 + 1862  = lim 10 + 10 − 1862 − 372 − 1862 − 10 + 1862 10 − 372 − 1862 = lim →0   = lim (10 − 372 − 186) = lim (10 − 372 − 186) = 10 − 372 →0  →0 →0 →0 The velocity of the rock when  =  is (10 − 372) ms (c) The rock will hit the surface when  = 0 ⇔ 10 − 1862 = 0 ⇔ (10 − 186) = 0 ⇔  = 0 or 186 = 10. The rock hits the surface when  = 10186 ≈ 54 s.  10   10  (d) The velocity of the rock when it hits the surface is  186 = 10 − 372 186 = 10 − 20 = −10 ms. 1 1 2 − ( + )2 − 2 2 ( + ) − () ( + )  2 ( + )2 2 − (2 + 2 + 2 ) = lim = lim = lim 15. () = lim →0 →0 →0 →0    2 ( + )2 = lim −(2 + 2 ) →0 2 ( + )2 So  (1) = = lim −(2 + ) →0 2 ( + )2 = lim −(2 + ) →0 2 ( + )2 = −2 −2 = 3 ms 2 · 2  −2 −2 −2 1 2 ms. = −2 ms, (2) = 3 = − ms, and (3) = 3 = − 13 2 4 3 27 16. (a) The average velocity between times  and  +  is 1 2 ( + ) − 6( + ) + 23 − ( + ) − () = 2 ( + ) −   1 2 1 2  2  − 6 + 23  +  + 12 2 − 6 − 6 + 23 − 12 2 + 6 − 23    1 1 2   −6   +  + 2  − 6 2 = =  + 12  − 6 fts =   (i) [4 8]:  = 4,  = 8 − 4 = 4, so the average velocity is 4 + 12 (4) − 6 = 0 fts. = 2 (ii) [6 8]:  = 6,  = 8 − 6 = 2, so the average velocity is 6 + 12 (2) − 6 = 1 fts. (iii) [8 10]:  = 8,  = 10 − 8 = 2, so the average velocity is 8 + 12 (2) − 6 = 3 fts. (iv) [8 12]:  = 8,  = 12 − 8 = 4, so the average velocity is 8 + 12 (4) − 6 = 4 fts. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 128 ¤ CHAPTER 2 LIMITS AND DERIVATIVES   ( + ) − () = lim  + 12  − 6 →0  =  − 6, so (8) = 2 fts. (b) () = lim →0 (c) 17.  0 (0) is the only negative value. The slope at  = 4 is smaller than the slope at  = 2 and both are smaller than the slope at  = −2. Thus, 0 (0)  0  0 (4)   0 (2)   0 (−2). 18. (a) On [20 60]: 700 − 300 400  (60) −  (20) = = = 10 60 − 20 40 40 (b) Pick any interval that has the same -value at its endpoints. [0 57] is such an interval since  (0) = 600 and  (57) = 600. (c) On [40 60]: 700 − 200 500  (60) −  (40) = = = 25 60 − 40 20 20 On [40 70]:  (70) −  (40) 900 − 200 700 = = = 23 13 70 − 40 30 30 Since 25  23 13 , the average rate of change on [40 60] is larger. (d) 200 − 400 −200  (40) −  (10) = = = −6 23 40 − 10 30 30 This value represents the slope of the line segment from (10 (10)) to (40 (40)). 19. (a) The tangent line at  = 50 appears to pass through the points (43 200) and (60 640), so  0 (50) ≈ 440 640 − 200 = ≈ 26. 60 − 43 17 (b) The tangent line at  = 10 is steeper than the tangent line at  = 30, so it is larger in magnitude, but less in numerical value, that is,  0 (10)   0 (30). (c) The slope of the tangent line at  = 60,  0 (60), is greater than the slope of the line through (40 (40)) and (80 (80)). So yes,  0 (60)  (80) −  (40) . 80 − 40 20. Since (5) = −3, the point (5 −3) is on the graph of . Since 0 (5) = 4, the slope of the tangent line at  = 5 is 4. Using the point-slope form of a line gives us  − (−3) = 4( − 5), or  = 4 − 23. 21. For the tangent line  = 4 − 5: when  = 2,  = 4(2) − 5 = 3 and its slope is 4 (the coefficient of ). At the point of tangency, these values are shared with the curve  =  (); that is,  (2) = 3 and  0 (2) = 4. 22. Since (4 3) is on  =  (),  (4) = 3. The slope of the tangent line between (0 2) and (4 3) is 14 , so  0 (4) = 14 . c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.7 DERIVATIVES AND RATES OF CHANGE 23. We begin by drawing a curve through the origin with a slope of 3 to satisfy  (0) = 0 and  0 (0) = 3. Since  0 (1) = 0, we will round off our figure so that there is a horizontal tangent directly over  = 1. Last, we make sure that the curve has a slope of −1 as we pass over  = 2. Two of the many possibilities are shown. 24. We begin by drawing a curve through the origin with a slope of 1 to satisfy (0) = 0 and  0 (0) = 1. We round off our figure at  = 1 to satisfy 0 (1) = 0, and then pass through (2 0) with slope −1 to satisfy (2) = 0 and  0 (2) = −1. We round the figure at  = 3 to satisfy  0 (3) = 0, and then pass through (4 0) with slope 1 to satisfy (4) = 0 and  0 (4) = 1 Finally we extend the curve on both ends to satisfy lim () = ∞ and lim () = −∞. →∞ →−∞ 25. We begin by drawing a curve through (0 1) with a slope of 1 to satisfy (0) = 1 and  0 (0) = 1. We round off our figure at  = −2 to satisfy  0 (−2) = 0. As  → −5+ ,  → ∞, so we draw a vertical asymptote at  = −5. As  → 5− ,  → 3, so we draw a dot at (5 3) [the dot could be open or closed]. 26. We begin by drawing an odd function (symmetric with respect to the origin) through the origin with slope −2 to satisfy  0 (0) = −2. Now draw a curve starting at  = 1 and increasing without bound as  → 2− since lim  () = ∞. Lastly, →2− reflect the last curve through the origin (rotate 180◦ ) since  is an odd function. 27. Using (4) with  () = 32 − 3 and  = 1,  (1 + ) −  (1) [3(1 + )2 − (1 + )3 ] − 2 = lim →0 →0    0 (1) = lim (3 + 6 + 32 ) − (1 + 3 + 32 + 3 ) − 2 3 − 3 (3 − 2 ) = lim = lim →0 →0 →0    = lim = lim (3 − 2 ) = 3 − 0 = 3 →0 Tangent line:  − 2 = 3( − 1) ⇔  − 2 = 3 − 3 ⇔  = 3 − 1 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° ¤ 129 130 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 28. Using (5) with () = 4 − 2 and  = 1,  0 (1) = lim →1 = lim →1 () − (1) (4 − 2) − (−1) 4 − 1 (2 + 1)(2 − 1) = lim = lim = lim →1 →1  − 1 →1 −1 −1 −1 (2 + 1)( + 1)( − 1) = lim [(2 + 1)( + 1)] = 2(2) = 4 →1 −1 Tangent line:  − (−1) = 4( − 1) ⇔  + 1 = 4 − 4 ⇔  = 4 − 5 29. (a) Using (4) with  () = 5(1 + 2 ) and the point (2 2), we have (b) 5(2 + ) −2  (2 + ) −  (2) 1 + (2 + )2  (2) = lim = lim →0 →0   0 5 + 10 = lim →0 2 + 4 + 5 −2  5 + 10 − 2(2 + 4 + 5) 2 + 4 + 5 = lim →0  −3 −22 − 3 (−2 − 3) −2 − 3 = lim = lim 2 = →0 (2 + 4 + 5) →0 (2 + 4 + 5) →0  + 4 + 5 5 = lim . So an equation of the tangent line at (2 2) is  − 2 = − 35 ( − 2) or  = − 35  + 16 5 30. (a) Using (4) with () = 42 − 3 , we have ( + ) − () [4( + )2 − ( + )3 ] − (42 − 3 ) = lim →0 →0   0 () = lim 42 + 8 + 42 − (3 + 32  + 32 + 3 ) − 42 + 3 →0  = lim 8 + 42 − 32  − 32 − 3 (8 + 4 − 32 − 3 − 2 ) = lim →0 →0   = lim = lim (8 + 4 − 32 − 3 − 2 ) = 8 − 32 →0 At the point (2 8), 0 (2) = 16 − 12 = 4, and an equation of the (b) tangent line is  − 8 = 4( − 2), or  = 4. At the point (3 9), 0 (3) = 24 − 27 = −3, and an equation of the tangent line is  − 9 = −3( − 3), or  = −3 + 18 31. Use (4) with  () = 32 − 4 + 1.  ( + ) − () [3( + )2 − 4( + ) + 1] − (32 − 4 + 1)] = lim →0 →0    0 () = lim 32 + 6 + 32 − 4 − 4 + 1 − 32 + 4 − 1 6 + 32 − 4 = lim →0 →0   = lim = lim →0 (6 + 3 − 4) = lim (6 + 3 − 4) = 6 − 4 →0  c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.7 DERIVATIVES AND RATES OF CHANGE 32. Use (4) with  () = 23 + .  0 () = lim →0  ( + ) −  () [2( + )3 + ( + )] − (23 + ) = lim →0   = lim 23 + 62  + 62 + 23 +  +  − 23 −  62  + 62 + 23 +  = lim →0   = lim (62 + 6 + 22 + 1) = lim (62 + 6 + 22 + 1) = 62 + 1 →0  →0 →0 33. Use (4) with  () = (2 + 1)( + 3). 2 + 1 2( + ) + 1 −  ( + ) −  () ( + ) + 3 +3 = lim  () = lim →0 →0   0 = lim (2 + 2 + 1)( + 3) − (2 + 1)( +  + 3) ( +  + 3)( + 3) = lim (22 + 6 + 2 + 6 +  + 3) − (22 + 2 + 6 +  +  + 3) ( +  + 3)( + 3) →0 →0 = lim 5 →0 ( +  + 3)( + 3) = lim 5 →0 ( +  + 3)( + 3) = 5 ( + 3)2 34. Use (4) with  () = −2 = 12 . 1 1 2 − ( + )2 −  ( + ) −  () ( + )2 2 2 ( + )2 = lim = lim  0 () = lim →0 →0 →0    2 − (2 + 2 + 2 ) −2 − 2 (−2 − ) = lim = lim 2 2 →0 →0 2 ( + )2 →0 2 ( + )2  ( + ) = lim = lim −2 −  →0 2 ( + )2 35. Use (4) with  () = = −2 −2 = 3 2 (2 )  √ 1 − 2.  √ 1 − 2( + ) − 1 − 2  ( + ) −  () = lim  () = lim →0 →0     √ √ 1 − 2( + ) − 1 − 2 1 − 2( + ) + 1 − 2 · = lim √ →0  1 − 2( + ) + 1 − 2  2 √ 2 1 − 2( + ) − 1 − 2 (1 − 2 − 2) − (1 − 2)   = lim   = lim √ √ →0 →0  1 − 2( + ) + 1 − 2  1 − 2( + ) + 1 − 2 0 = lim →0 −2 −2   = lim  √ √ →0 1 − 2( + ) + 1 − 2  1 − 2( + ) + 1 − 2 −2 −1 −2 √ = √ = √ = √ 1 − 2 + 1 − 2 2 1 − 2 1 − 2 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° ¤ 131 132 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 4 . 1− 36. Use (4) with  () = √ 4 4  −√ 1 − 1 − ( + )  ( + ) −  ()  0 () = lim = lim →0 →0   √ √ 1−− 1−− √ √ √ √ 1−− 1−− 1−− 1− = 4 lim √ = 4 lim √ →0 →0  1 −  −   1− √ √ √ √ √ √ 1−− 1−− 1−+ 1−− ( 1 − )2 − ( 1 −  − )2 √ √ = 4 lim √ ·√ = 4 lim √ √ √ √ →0  1 −  −  →0  1 −  −  1− 1−+ 1−− 1 − ( 1 −  + 1 −  − ) (1 − ) − (1 −  − )  √ √ √ = 4 lim √ √ √ √ √ →0  1 −  −  1 −  −  1 − ( 1 −  + 1 −  − ) 1 − ( 1 −  + 1 −  − ) = 4 lim →0  1 1 √ √ √ √ =4· √ = 4 lim √ √ √ →0 1 −  1 − ( 1 −  + 1 − ) 1 −  −  1 − ( 1 −  + 1 −  − ) = 2 4 2 √ = = 1 (1 − )12 (1 − ) (1 − )32 (1 − )(2 1 − ) √ √ 9+−3 =  0 (9), where  () =  and  = 9. →0  37. By (4), lim −2+ − −2 =  0 (−2), where () =  and  = −2. →0  38. By (4), lim 6 − 64 =  0 (2), where  () = 6 and  = 2. →2  − 2 39. By Equation 5, lim 1 −4 1 1  =  0 (4), where  () = and  = . 40. By Equation 5, lim 1 →14  4 − 4 41. By (4), lim →0 cos( + ) + 1 =  0 (), where  () = cos  and  = .  Or: By (4), lim →0 cos( + ) + 1 =  0 (0), where  () = cos( + ) and  = 0.    sin  − 12  0  =  , where  () = sin  and  = .  →6  − 6 6 6 42. By Equation 5, lim     80(4 + ) − 6(4 + )2 − 80(4) − 6(4)2  (4 + ) −  (4) = lim 43. (4) =  (4) = lim →0 →0   0 = lim (320 + 80 − 96 − 48 − 62 ) − (320 − 96) 32 − 62 = lim →0   = lim (32 − 6) = lim (32 − 6) = 32 m/s →0  →0 →0 The speed when  = 4 is |32| = 32 ms. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.7  (4 + ) −  (4) = lim 44. (4) =  (4) = lim →0 →0  0  10 + 45 4++1   DERIVATIVES AND RATES OF CHANGE   45 − 10 + 4+1 ¤ 133 45 −9 5 = lim +  →0  45 − 9(5 + ) 9 −9 −9 = lim = lim = − m/s. →0 (5 + ) →0 5 +  (5 + ) 5   The speed when  = 4 is − 95  = 95 ms. = lim →0 45. The sketch shows the graph for a room temperature of 72◦ and a refrigerator temperature of 38◦ . The initial rate of change is greater in magnitude than the rate of change after an hour. 46. The slope of the tangent (that is, the rate of change of temperature with respect to time) at  = 1 h seems to be about 47. (a) (i) [10 20]: 75 − 168 ≈ −07 ◦ Fmin. 132 − 0 018 − 033 mg/mL (2) − (1) = = −015 2−1 1 h (ii) [15 20]: 018 − 024 −006 mg/mL (2) − (15) = = = −012 2 − 15 05 05 h (iii) [20 25]: 012 − 018 −006 mg/mL (25) − (2) = = = −012 25 − 2 05 05 h (iv) [20 30]: 007 − 018 mg/mL (3) − (2) = = −011 3−2 1 h (b) We estimate the instantaneous rate of change at  = 2 by averaging the average rates of change for [15 20] and [20 25]: mg/mL −012 + (−012) = −012 . After 2 hours, the BAC is decreasing at a rate of 012 (mgmL)h. 2 h 48. (a) (i) [2006 2008]: (ii) [2008 2010]: 16,680 − 12,440 4240 (2008) − (2006) = = = 2120 locationsyear 2008 − 2006 2 2 16,858 − 16,680 178 (2010) − (2008) = = = 89 locationsyear. 2010 − 2008 2 2 The rate of growth decreased over the period from 2006 to 2010. (b) [2010 2012]: (2012) − (2010) 18,066 − 16,858 1208 = = = 604 locationsyear. 2012 − 2010 2 2 Using that value and the value from part (a)(ii), we have 693 89 + 604 = = 3465 locationsyear. 2 2 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 134 ¤ CHAPTER 2 LIMITS AND DERIVATIVES (c) The tangent segment has endpoints (2008 16,250) and (2012 17,500). An estimate of the instantaneous rate of growth in 2010 is 17,500 − 16,250 1250 = = 3125 locations/year. 2012 − 2008 4 49. (a) [1990 2005]: 17,544 84,077 − 66,533 = = 11696 thousands of barrels per day per year. This means that oil 2005 − 1990 15 consumption rose by an average of 11696 thousands of barrels per day each year from 1990 to 2005. (b) [1995 2000]: [2000 2005]: 6685 76,784 − 70,099 = = 1337 2000 − 1995 5 7293 84,077 − 76,784 = = 14586 2005 − 2000 5 An estimate of the instantaneous rate of change in 2000 is 12 (1337 + 14586) = 13978 thousands of barrels per day per year. 50. (a) (i) [4 11]: 94 − 53 −436 RNA copiesmL  (11) −  (4) = = ≈ −623 11 − 4 7 7 day (ii) [8 11]: 94 − 18 −86 RNA copiesmL  (11) −  (8) = = ≈ −287 11 − 8 3 3 day (iii) [11 15]: 52 − 94 −42 RNA copiesmL  (15) −  (11) = = = −105 15 − 11 4 4 day (iv) [11 22]: 36 − 94 −58 RNA copiesmL  (22) −  (11) = = ≈ −053 22 − 11 11 11 day (b) An estimate of  0 (11) is the average of the answers from part (a)(ii) and (iii).  0 (11) ≈ 12 [−287 + (−105)] = −196 RNA copiesmL . day  0 (11) measures the instantaneous rate of change of patient 303’s viral load 11 days after ABT-538 treatment began. 51. (a) (i) (105) − (100) 660125 − 6500 ∆ = = = $2025unit. ∆ 105 − 100 5 (ii) (101) − (100) 652005 − 6500 ∆ = = = $2005unit. ∆ 101 − 100 1 (b)   5000 + 10(100 + ) + 005(100 + )2 − 6500 (100 + ) − (100) 20 + 0052 = =    = 20 + 005,  6= 0 So the instantaneous rate of change is lim →0 (100 + ) − (100) = lim (20 + 005) = $20unit. →0  c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.7 DERIVATIVES AND RATES OF CHANGE ¤ 135 2 2   +  52. ∆ =  ( + ) −  () = 100,000 1 − − 100,000 1 − 60 60      2 ( + )  2  2 2 + + − 1− + = 100,000 − + + = 100,000 1 − 30 3600 30 3600 30 3600 3600 = 250 100,000  (−120 + 2 + ) =  (−120 + 2 + ) 3600 9 ( − 60) galmin. Dividing ∆ by  and then letting  → 0, we see that the instantaneous rate of change is 500 9  Flow rate (galmin) Water remaining  () (gal) 0 −33333 100 000 −22222 44 4444 −11111 11 1111 10 20 30 40 50 60 −27777 69 4444 −16666 25 000 − 5555 2 7777 0 0 The magnitude of the flow rate is greatest at the beginning and gradually decreases to 0. 53. (a)  0 () is the rate of change of the production cost with respect to the number of ounces of gold produced. Its units are dollars per ounce. (b) After 800 ounces of gold have been produced, the rate at which the production cost is increasing is $17ounce. So the cost of producing the 800th (or 801st) ounce is about $17. (c) In the short term, the values of  0 () will decrease because more efficient use is made of start-up costs as  increases. But eventually  0 () might increase due to large-scale operations. 54. (a)  0 (5) is the rate of growth of the bacteria population when  = 5 hours. Its units are bacteria per hour. (b) With unlimited space and nutrients,  0 should increase as  increases; so  0 (5)   0 (10). If the supply of nutrients is limited, the growth rate slows down at some point in time, and the opposite may be true. 55. (a)  0 (58) is the rate at which the daily heating cost changes with respect to temperature when the outside temperature is 58 ◦ F. The units are dollars ◦ F. (b) If the outside temperature increases, the building should require less heating, so we would expect  0 (58) to be negative. 56. (a)  0 (8) is the rate of change of the quantity of coffee sold with respect to the price per pound when the price is $8 per pound. The units for  0 (8) are pounds(dollarspound). (b)  0 (8) is negative since the quantity of coffee sold will decrease as the price charged for it increases. People are generally less willing to buy a product when its price increases. 57. (a)  0 ( ) is the rate at which the oxygen solubility changes with respect to the water temperature. Its units are (mgL)◦ C. (b) For  = 16◦ C, it appears that the tangent line to the curve goes through the points (0 14) and (32 6). So  0 (16) ≈ 8 6 − 14 =− = −025 (mgL)◦ C. This means that as the temperature increases past 16◦ C, the oxygen 32 − 0 32 solubility is decreasing at a rate of 025 (mgL)◦ C. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 136 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 58. (a)  0 ( ) is the rate of change of the maximum sustainable speed of Coho salmon with respect to the temperature. Its units are (cms)◦ C. (b) For  = 15◦ C, it appears the tangent line to the curve goes through the points (10 25) and (20 32). So  0 (15) ≈ 32 − 25 = 07 (cms)◦ C. This tells us that at  = 15◦ C, the maximum sustainable speed of Coho salmon is 20 − 10 changing at a rate of 0.7 (cms)◦ C. In a similar fashion for  = 25◦ C, we can use the points (20 35) and (25 25) to obtain  0 (25) ≈ 25 − 35 = −2 (cms)◦ C. As it gets warmer than 20◦ C, the maximum sustainable speed decreases 25 − 20 rapidly. 59. Since  () =  sin(1) when  6= 0 and  (0) = 0, we have  0 (0) = lim →0  (0 + ) −  (0)  sin(1) − 0 = lim = lim sin(1). This limit does not exist since sin(1) takes the →0 →0   values −1 and 1 on any interval containing 0. (Compare with Example 2.2.4.) 60. Since () = 2 sin(1) when  6= 0 and  (0) = 0, we have 1  (0 + ) −  (0) 2 sin(1) − 0 = lim = lim  sin(1). Since −1 ≤ sin ≤ 1, we have →0 →0 →0     0 (0) = lim 1 1 ≤ || ⇒ − || ≤  sin ≤ ||. Because lim (− ||) = 0 and lim || = 0, we know that →0 →0     1 = 0 by the Squeeze Theorem. Thus,  0 (0) = 0. lim  sin →0  − || ≤ || sin 61. (a) The slope at the origin appears to be 1. (b) The slope at the origin still appears to be 1. (c) Yes, the slope at the origin now appears to be 0. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.8 THE DERIVATIVE AS A FUNCTION ¤ 137 2.8 The Derivative as a Function 1. It appears that  is an odd function, so  0 will be an even function—that is,  0 (−) =  0 (). (a)  0 (−3) ≈ −02 (b)  0 (−2) ≈ 0 (c)  0 (−1) ≈ 1 (d)  0 (0) ≈ 2 (e)  0 (1) ≈ 1 (f)  0 (2) ≈ 0 (g)  0 (3) ≈ −02 2. Your answers may vary depending on your estimates. (a) Note: By estimating the slopes of tangent lines on the graph of  , it appears that  0 (0) ≈ 6. (b)  0 (1) ≈ 0 (c)  0 (2) ≈ −15 (d)  0 (3) ≈ −13 (e)  0 (4) ≈ −08 (f)  0 (5) ≈ −03 (g)  0 (6) ≈ 0 (h)  0 (7) ≈ 02 3. (a)0 = II, since from left to right, the slopes of the tangents to graph (a) start out negative, become 0, then positive, then 0, then negative again. The actual function values in graph II follow the same pattern. (b) = IV, since from left to right, the slopes of the tangents to graph (b) start out at a fixed positive quantity, then suddenly 0 become negative, then positive again. The discontinuities in graph IV indicate sudden changes in the slopes of the tangents. (c) = I, since the slopes of the tangents to graph (c) are negative for   0 and positive for   0, as are the function values of 0 graph I. (d) = III, since from left to right, the slopes of the tangents to graph (d) are positive, then 0, then negative, then 0, then 0 positive, then 0, then negative again, and the function values in graph III follow the same pattern. Hints for Exercises 4 –11: First plot -intercepts on the graph of  0 for any horizontal tangents on the graph of  . Look for any corners on the graph of  — there will be a discontinuity on the graph of  0 . On any interval where  has a tangent with positive (or negative) slope, the graph of  0 will be positive (or negative). If the graph of the function is linear, the graph of  0 will be a horizontal line. 4. 5. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 138 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 6. 7. 8. 9. 10. 11. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.8 THE DERIVATIVE AS A FUNCTION 12. The slopes of the tangent lines on the graph of  =  () are always positive, so the -values of  =  0() are always positive. These values start out relatively small and keep increasing, reaching a maximum at about  = 6. Then the -values of  =  0() decrease and get close to zero. The graph of  0 tells us that the yeast culture grows most rapidly after 6 hours and then the growth rate declines. 13. (a)  0 () is the instantaneous rate of change of percentage of full capacity with respect to elapsed time in hours. (b) The graph of  0 () tells us that the rate of change of percentage of full capacity is decreasing and approaching 0. 14. (a)  0 () is the instantaneous rate of change of fuel economy with respect to speed. (b) Graphs will vary depending on estimates of  0 , but will change from positive to negative at about  = 50. (c) To save on gas, drive at the speed where  is a maximum and  0 is 0, which is about 50 mi h. 15. It appears that there are horizontal tangents on the graph of  for  = 1963 and  = 1971. Thus, there are zeros for those values of  on the graph of  0 . The derivative is negative for the years 1963 to 1971. 16. See Figure 3.3.1. 17. The slope at 0 appears to be 1 and the slope at 1 appears to be 27. As  decreases, the slope gets closer to 0. Since the graphs are so similar, we might guess that  0 () =  . c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° ¤ 139 140 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 18. As  increases toward 1,  0 () decreases from very large numbers to 1. As  becomes large,  0 () gets closer to 0. As a guess,  0 () = 12 or  0 () = 1 makes sense.   19. (a) By zooming in, we estimate that  0 (0) = 0,  0 12 = 1,  0 (1) = 2, and  0 (2) = 4.   (b) By symmetry,  0 (−) = − 0 (). So  0 − 12 = −1,  0 (−1) = −2, and  0 (−2) = −4. (c) It appears that  0 () is twice the value of , so we guess that  0 () = 2.  ( + ) −  () ( + )2 − 2 = lim →0 →0    2   + 2 + 2 − 2 2 + 2 (2 + ) = lim = lim = lim = lim (2 + ) = 2 →0 →0 →0 →0    (d)  0 () = lim   20. (a) By zooming in, we estimate that  0 (0) = 0,  0 12 ≈ 075,  0 (1) ≈ 3,  0 (2) ≈ 12, and  0 (3) ≈ 27.   (b) By symmetry,  0 (−) =  0 (). So  0 − 12 ≈ 075,  0 (−1) ≈ 3,  0 (−2) ≈ 12, and  0 (−3) ≈ 27. (c) (d) Since  0 (0) = 0, it appears that  0 may have the form  0 () = 2 . Using  0 (1) = 3, we have  = 3, so  0 () = 32 .  ( + ) −  () ( + )3 − 3 (3 + 32  + 32 + 3 ) − 3 = lim = lim →0 →0 →0    (e)  0 () = lim 32  + 32 + 3 (32 + 3 + 2 ) = lim = lim (32 + 3 + 2 ) = 32 →0 →0 →0   = lim c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.8 21.  0 () = lim →0 = lim →0 THE DERIVATIVE AS A FUNCTION ¤ ( + ) −  () [3( + ) − 8] − (3 − 8) 3 + 3 − 8 − 3 + 8 = lim = lim →0 →0    3 = lim 3 = 3 →0  Domain of  = domain of  0 = R. 22.  0 () = lim →0 ( + ) −  () [( + ) + ] − ( + )  +  +  −  −  = lim = lim →0 →0     = lim  =  →0  Domain of  = domain of  0 = R. = lim →0     25( + )2 + 6( + ) − 252 + 6  ( + ) − () = lim 23.  () = lim →0 →0   0 = lim 25(2 + 2 + 2 ) + 6 + 6 − 252 − 6 252 + 5 + 252 + 6 − 252 = lim →0   = lim  (5 + 25 + 6) 5 + 252 + 6 = lim = lim (5 + 25 + 6) →0 →0   →0 →0 = 5 + 6 Domain of  = domain of  0 = R.   4 + 8( + ) − 5( + )2 − (4 + 8 − 52 ) ( + ) −  () = lim →0 →0   24.  0 () = lim = lim 4 + 8 + 8 − 5(2 + 2 + 2 ) − 4 − 8 + 52 8 − 52 − 10 − 52 + 52 = lim →0   = lim 8 − 10 − 52 (8 − 10 − 5) = lim = lim (8 − 10 − 5) →0 →0   →0 →0 = 8 − 10 Domain of  = domain of  0 = R. 25.  0 () = lim →0 = lim →0 ( + ) −  () [( + )2 − 2( + )3 ] − (2 − 23 ) = lim →0   2 + 2 + 2 − 23 − 62  − 62 − 23 − 2 + 23  2 + 2 − 62  − 62 − 23 (2 +  − 62 − 6 − 22 ) = lim →0   2 2 2 = lim (2 +  − 6 − 6 − 2 ) = 2 − 6 = lim →0 →0 Domain of  = domain of  0 = R. √ √ − + 1 1 √ √ √ √  √ √ √ −√ ( + ) − () − + + + + +   √ ·√ √ √ = lim = lim = lim 26. 0 () = lim →0 →0 →0 →0     +  + + = lim →0   − ( + ) − −1 √ √ √ √ √ √ √ √  = lim √ √  = lim √ √  →0  →0 +  + + +  + + +  + + −1 1 −1 √  =  √  = − 32 = √ √ √ 2   +   2  Domain of  = domain of 0 = (0 ∞). c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 141 142 ¤ CHAPTER 2 LIMITS AND DERIVATIVES ( + ) − () = lim 27.  () = lim →0 →0  0 = lim →0    √ √ 9 − ( + ) − 9 −  9 − ( + ) + 9 −   √  9 − ( + ) + 9 −  [9 − ( + )] − (9 − ) −   = lim   √ √ →0  9 − ( + ) + 9 −   9 − ( + ) + 9 −  −1 −1 = lim  = √ √ →0 2 9− 9 − ( + ) + 9 −  Domain of  = (−∞ 9], domain of 0 = (−∞ 9). ( + )2 − 1 2 − 1 − ( + ) −  () 2( + ) − 3 2 − 3 = lim 28.  0 () = lim →0 →0   [( + )2 − 1](2 − 3) − [2( + ) − 3](2 − 1) [2( + ) − 3](2 − 3) = lim →0  (2 + 2 + 2 − 1)(2 − 3) − (2 + 2 − 3)(2 − 1) →0 [2( + ) − 3](2 − 3) = lim (23 + 42  + 22 − 2 − 32 − 6 − 32 + 3) − (23 + 22  − 32 − 2 − 2 + 3) →0 (2 + 2 − 3)(2 − 3) = lim 42  + 22 − 6 − 32 − 22  + 2 (22 + 2 − 6 − 3 + 2) = lim →0 →0 (2 + 2 − 3)(2 − 3) (2 + 2 − 3)(2 − 3) = lim 22 − 6 + 2 22 + 2 − 6 − 3 + 2 = →0 (2 + 2 − 3)(2 − 3) (2 − 3)2 = lim Domain of  = domain of  0 = (−∞ 32 ) ∪ ( 32  ∞). 1 − 2 1 − 2( + ) − ( + ) − () 3 + ( + ) 3+ = lim 29.  () = lim →0 →0   0 [1 − 2( + )](3 + ) − [3 + ( + )](1 − 2) [3 + ( + )](3 + ) = lim →0  = lim →0 = lim 3 +  − 6 − 22 − 6 − 2 − (3 − 6 +  − 22 +  − 2) −6 −  = lim →0 (3 +  + )(3 + ) [3 + ( + )](3 + ) −7 →0 (3 +  + )(3 + ) = lim −7 →0 (3 +  + )(3 + ) = −7 (3 + )2 Domain of  = domain of 0 = (−∞ −3) ∪ (−3 ∞). ( + ) −  () ( + )32 − 32 [( + )32 − 32 ][( + )32 + 32 ] = lim = lim →0 →0    [( + )32 + 32 ]    32 + 3 + 2 ( + )3 − 3 3 + 32  + 32 + 3 − 3 = lim = lim = lim →0 [( + )32 + 32 ] →0 →0 [( + )32 + 32 ] [( + )32 + 32 ] 30.  0 () = lim →0 32 32 + 3 + 2 = 32 = 32 12 32 32 →0 ( + ) + 2 = lim Domain of  = domain of  0 = [0 ∞). Strictly speaking, the domain of  0 is (0 ∞) because the limit that defines  0 (0) does c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.8 THE DERIVATIVE AS A FUNCTION ¤ 143 not exist (as a two-sided limit). But the right-hand derivative (in the sense of Exercise 64) does exist at 0, so in that sense one could regard the domain of  0 to be [0 ∞).  4   + 43  + 62 2 + 43 + 4 − 4 ( + ) −  () ( + )4 − 4 = lim = lim 31.  () = lim →0 →0 →0     3  43  + 62 2 + 43 + 4 = lim = lim 4 + 62  + 42 + 3 = 43 →0 →0  0 Domain of  = domain of  0 = R. 32. (a) (b) Note that the third graph in part (a) has small negative values for its slope,  0 ; but as  → 6− ,  0 → −∞. See the graph in part (d). ( + ) −  ()     √ √ 6 − ( + ) − 6 −  6 − ( + ) + 6 −   = lim √ →0  6 − ( + ) + 6 −  (c)  0 () = lim →0 = lim →0 (d) [6 − ( + )] − (6 − ) −   = lim √  √ √ →0  6 −  − + 6−  6 − ( + ) + 6 −  −1 −1 = √ = lim √ √ →0 2 6− 6−−+ 6− Domain of  = (−∞ 6], domain of  0 = (−∞ 6). ( + ) −  () [( + )4 + 2( + )] − (4 + 2) = lim →0 →0   33. (a)  0 () = lim 4 + 43  + 62 2 + 43 + 4 + 2 + 2 − 4 − 2 →0  = lim = lim →0 43  + 62 2 + 43 + 4 + 2 (43 + 62  + 42 + 3 + 2) = lim →0   = lim (43 + 62  + 42 + 3 + 2) = 43 + 2 →0 (b) Notice that  0 () = 0 when  has a horizontal tangent,  0 () is positive when the tangents have positive slope, and  0 () is negative when the tangents have negative slope. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 144 ¤ CHAPTER 2 LIMITS AND DERIVATIVES  ( + ) −  () [( + ) + 1( + )] − ( + 1) = lim = lim 34. (a)  0 () = lim →0 →0 →0   2 + 1 ( + )2 + 1 − +   [( + )2 + 1] − ( + )(2 + 1) (3 + 22 + 2 + ) − (3 +  + 2 + ) = lim →0 →0 ( + ) ( + ) = lim 2 − 1 1 2 + 2 −  (2 +  − 1) 2 +  − 1 = lim = lim = , or 1 − 2 →0 →0 →0 ( + ) ( + ) ( + ) 2  = lim (b) Notice that  0 () = 0 when  has a horizontal tangent,  0 () is positive when the tangents have positive slope, and  0 () is negative when the tangents have negative slope. Both functions are discontinuous at  = 0. 35. (a)  0 () is the rate at which the unemployment rate is changing with respect to time. Its units are percent unemployed per year. (b) To find  0 (), we use lim →0 For 2003:  0 (2003) ≈ ( + ) −  ()  ( + ) − () ≈ for small values of .   55 − 60  (2004) − (2003) = = −05 2004 − 2003 1 For 2004: We estimate  0 (2004) by using  = −1 and  = 1, and then average the two results to obtain a final estimate.  = −1 ⇒  0 (2004) ≈  = 1 ⇒  0 (2004) ≈ 60 − 55  (2003) −  (2004) = = −05; 2003 − 2004 −1  (2005) −  (2004) 51 − 55 = = −04. 2005 − 2004 1 So we estimate that  0 (2004) ≈ 12 [−05 + (−04)] = −045. Other values for  0 () are calculated in a similar fashion.  2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 0 −050 −045 −045 −025 060 235 190 −020 −075 −080  () 36. (a)  0 () is the rate at which the number of minimally invasive cosmetic surgery procedures performed in the United States is changing with respect to time. Its units are thousands of surgeries per year. (b) To find  0 (), we use lim →0 For 2000:  0 (2000) ≈ ( + ) − () ( + ) − () ≈ for small values of .   4897 − 5500 (2002) − (2000) = = −3015 2002 − 2000 2 For 2002: We estimate  0 (2002) by using  = −2 and  = 2, and then average the two results to obtain a final estimate.  = −2 ⇒  0 (2002) ≈  = 2 ⇒  0 (2002) ≈ 5500 − 4897 (2000) − (2002) = = −3015 2000 − 2002 −2 7470 − 4897 (2004) − (2002) = = 12865 2004 − 2002 2 So we estimate that  0 (2002) ≈ 12 [−3015 + 12865] = 4925.  2000 2002 2004 2006 2008 2010 2012 0 −3015 4925 106025 85675 60575 5345 737  () c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.8 (c) THE DERIVATIVE AS A FUNCTION ¤ 145 (d) We could get more accurate values for  0 () by obtaining data for more values of . 37. As in Exercise 35, we use one-sided difference quotients for the first and last values, and average two difference quotients for all other values.  14 21 28 35 42 49 () 41 54 64 72 78 83  0 () 13 7 23 14 18 14 14 14 11 14 5 7 38. As in Exercise 35, we use one-sided difference quotients for the first and last values, and average two difference quotients for all other values. The units for  0 () are grams per degree (g◦ C).  155 177 200 224 244  () 372 310 198 97 0 −98 −282 −387 −453 −673 −975  () 39. (a)  is the rate at which the percentage of the city’s electrical power produced by solar panels changes with respect to time , measured in percentage points per year. (b) 2 years after January 1, 2000 (January 1, 2002), the percentage of electrical power produced by solar panels was increasing at a rate of 3.5 percentage points per year. 40.  is the rate at which the number of people who travel by car to another state for a vacation changes with respect to the price of gasoline. If the price of gasoline goes up, we would expect fewer people to travel, so we would expect  to be negative. 41.  is not differentiable at  = −4, because the graph has a corner there, and at  = 0, because there is a discontinuity there. 42.  is not differentiable at  = −1, because there is a discontinuity there, and at  = 2, because the graph has a corner there. 43.  is not differentiable at  = 1, because  is not defined there, and at  = 5, because the graph has a vertical tangent there. 44.  is not differentiable at  = −2 and  = 3, because the graph has corners there, and at  = 1, because there is a discontinuity there. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 146 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 45. As we zoom in toward (−1 0), the curve appears more and more like a straight line, so  () =  +  || is differentiable at  = −1. But no matter how much we zoom in toward the origin, the curve doesn’t straighten out—we can’t eliminate the sharp point (a cusp). So  is not differentiable at  = 0. 46. As we zoom in toward (0 1), the curve appears more and more like a straight line, so () = (2 − 1)23 is differentiable at  = 0. But no matter how much we zoom in toward (1 0) or (−1 0), the curve doesn’t straighten out—we can’t eliminate the sharp point (a cusp). So  is not differentiable at  = ±1. 47. Call the curve with the positive -intercept  and the other curve . Notice that  has a maximum (horizontal tangent) at  = 0, but  6= 0, so  cannot be the derivative of . Also notice that where  is positive,  is increasing. Thus,  =  and  =  0 . Now  0 (−1) is negative since  0 is below the -axis there and  00 (1) is positive since  is concave upward at  = 1. Therefore,  00 (1) is greater than  0 (−1). 48. Call the curve with the smallest positive -intercept  and the other curve . Notice that where  is positive in the first quadrant,  is increasing. Thus,  =  and  =  0 . Now  0 (−1) is positive since  0 is above the -axis there and  00 (1) appears to be zero since  has an inflection point at  = 1. Therefore,  0 (1) is greater than  00 (−1). 49.  =  ,  =  0 ,  =  00 . We can see this because where  has a horizontal tangent,  = 0, and where  has a horizontal tangent,  = 0. We can immediately see that  can be neither  nor  0 , since at the points where  has a horizontal tangent, neither  nor  is equal to 0. 50. Where  has horizontal tangents, only  is 0, so 0 = .  has negative tangents for   0 and  is the only graph that is negative for   0, so 0 = .  has positive tangents on R (except at  = 0), and the only graph that is positive on the same domain is , so 0 = . We conclude that  =  ,  =  0 ,  =  00 , and  =  000 . 51. We can immediately see that  is the graph of the acceleration function, since at the points where  has a horizontal tangent, neither  nor  is equal to 0. Next, we note that  = 0 at the point where  has a horizontal tangent, so  must be the graph of the velocity function, and hence, 0 = . We conclude that  is the graph of the position function. 52.  must be the jerk since none of the graphs are 0 at its high and low points.  is 0 where  has a maximum, so 0 = .  is 0 where  has a maximum, so 0 = . We conclude that  is the position function,  is the velocity,  is the acceleration, and  is the jerk. ( + ) −  () [3( + )2 + 2( + ) + 1] − (32 + 2 + 1) = lim →0 →0   53.  0 () = lim (32 + 6 + 32 + 2 + 2 + 1) − (32 + 2 + 1) 6 + 32 + 2 = lim →0 →0   = lim = lim →0 (6 + 3 + 2) = lim (6 + 3 + 2) = 6 + 2 →0  c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.8 THE DERIVATIVE AS A FUNCTION ¤  0 ( + ) −  0 () [6( + ) + 2] − (6 + 2) (6 + 6 + 2) − (6 + 2) = lim = lim →0 →0     00 () = lim →0 6 = lim 6 = 6 →0  = lim →0 We see from the graph that our answers are reasonable because the graph of  0 is that of a linear function and the graph of  00 is that of a constant function. ( + ) −  () [( + )3 − 3( + )] − (3 − 3) = lim →0 →0   (3 + 32  + 32 + 3 − 3 − 3) − (3 − 3) 32  + 32 + 3 − 3 = lim = lim →0 →0   54.  0 () = lim = lim →0 (32 + 3 + 2 − 3) = lim (32 + 3 + 2 − 3) = 32 − 3 →0   0 ( + ) −  0 () [3( + )2 − 3] − (32 − 3) (32 + 6 + 32 − 3) − (32 − 3) = lim = lim →0 →0 →0     00 () = lim 6 + 32 (6 + 3) = lim = lim (6 + 3) = 6 →0 →0 →0   = lim We see from the graph that our answers are reasonable because the graph of  0 is that of an even function ( is an odd function) and the graph of  00 is that of an odd function. Furthermore,  0 = 0 when  has a horizontal tangent and  00 = 0 when  0 has a horizontal tangent.   2( + )2 − ( + )3 − (22 − 3 )  ( + ) −  () = lim 55.  () = lim →0 →0   0 (4 + 2 − 32 − 3 − 2 ) = lim (4 + 2 − 32 − 3 − 2 ) = 4 − 32 →0 →0    4( + ) − 3( + )2 − (4 − 32 )  0 ( + ) −  0 () (4 − 6 − 3) 00 = lim = lim  () = lim →0 →0 →0    = lim = lim (4 − 6 − 3) = 4 − 6 →0  00 ( + ) −  00 () [4 − 6( + )] − (4 − 6) −6 = lim = lim = lim (−6) = −6 →0 →0 →0  →0    000 () = lim  000 ( + ) −  000 () −6 − (−6) 0 = lim = lim = lim (0) = 0 →0 →0 →0  →0    (4) () = lim The graphs are consistent with the geometric interpretations of the derivatives because  0 has zeros where  has a local minimum and a local maximum,  00 has a zero where  0 has a local maximum, and  000 is a constant function equal to the slope of  00 . c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 147 148 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 56. (a) Since we estimate the velocity to be a maximum at  = 10, the acceleration is 0 at  = 10. (b) Drawing a tangent line at  = 10 on the graph of ,  appears to decrease by 10 fts2 over a period of 20 s. So at  = 10 s, the jerk is approximately −1020 = −05 (fts2 )s or fts3 . 57. (a) Note that we have factored  −  as the difference of two cubes in the third step.  () −  () 13 − 13 13 − 13 = lim = lim 13 → → → ( − − − 13 )(23 + 13 13 + 23 )  0() = lim 1 = lim → 23 + 13 13 + 23 = 1 or 13 −23 323 √ 3 (0 + ) − (0) −0 1 = lim = lim 23 . This function increases without bound, so the limit does not →0 →0 →0    (b)  0(0) = lim exist, and therefore  0(0) does not exist. (c) lim | 0 ()| = lim 1 →0 323 →0 = ∞ and  is continuous at  = 0 (root function), so  has a vertical tangent at  = 0. () − (0) 23 − 0 1 = lim = lim 13 , which does not exist. →0 →0 →0  −0  58. (a)  0(0) = lim () − () 23 − 23 (13 − 13 )(13 + 13 ) = lim = lim 13 → → → ( − − − 13 )(23 + 13 13 + 23 ) (b)  0() = lim 13 + 13 = lim → 23 + 13 13 + 23 = 213 2 = 13 or 23 −13 323 3 (c) () = 23 is continuous at  = 0 and lim |0()| = lim →0 2 →0 3 ||13 (d) = ∞. This shows that  has a vertical tangent line at  = 0. 59.  () = | − 6| =  −6 −( − 6) if  − 6  0 So the right-hand limit is lim →6+ is lim →6− if  − 6 ≥ 6 =   − 6 if  ≥ 6 6 −  if   6  () −  (6) | − 6| − 0 −6 = lim = lim = lim 1 = 1, and the left-hand limit −6 −6 →6+ →6+  − 6 →6+ () − (6) | − 6| − 0 6− = lim = lim = lim (−1) = −1. Since these limits are not equal, −6 −6 →6− →6−  − 6 →6− c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° SECTION 2.8 THE DERIVATIVE AS A FUNCTION  () −  (6) does not exist and  is not differentiable at 6. −6  1 if   6 0 0 However, a formula for  is  () = −1 if   6  0 (6) = lim →6 Another way of writing the formula is  0 () = −6 . | − 6| 60.  () = [[]] is not continuous at any integer , so  is not differentiable at  by the contrapositive of Theorem 4. If  is not an integer, then  is constant on an open interval containing , so  0() = 0. Thus,  0() = 0,  not an integer.  2  if  ≥ 0 61. (a)  () =  || = 2 − if   0 (b) Since  () = 2 for  ≥ 0, we have  0 () = 2 for   0. [See Exercise 19(d).] Similarly, since  () = −2 for   0, we have  0 () = −2 for   0. At  = 0, we have  0 (0) = lim →0  () −  (0)  || = lim = lim || = 0 →0  →0 −0 So  is differentiable at 0. Thus,  is differentiable for all . 0  2 if  ≥ 0 (c) From part (b), we have  () = 62. (a) || =  2 if  ≥ 0 −2 if   0  = 2 ||. if  ≥ 0  − if   0  so () =  + || = 0 if   0 . Graph the line  = 2 for  ≥ 0 and graph  = 0 (the x-axis) for   0. (b)  is not differentiable at  = 0 because the graph has a corner there, but is differentiable at all other values; that is,  is differentiable on (−∞ 0) ∪ (0 ∞). (c) () =  2 0 if  ≥ 0 if   0 0 ⇒  () =  2 if   0 0 if   0 Another way of writing the formula is 0 () = 1 + sgn  for  6= 0. 63. (a) If  is even, then  0 (−) = lim →0 = lim →0  (− + ) −  (−)  [−( − )] −  (−) = lim →0    ( − ) − ()  ( − ) −  () = − lim →0  − = − lim ∆→0 [let ∆ = −]  ( + ∆) −  () = − 0 () ∆ Therefore,  0 is odd. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° ¤ 149 150 ¤ CHAPTER 2 LIMITS AND DERIVATIVES (b) If  is odd, then  0 (−) = lim →0 = lim →0 = lim  (− + ) −  (−)  [−( − )] −  (−) = lim →0   −( − ) +  ()  ( − ) − () = lim →0  − ∆→0 [let ∆ = −]  ( + ∆) −  () =  0 () ∆ Therefore,  0 is even.  (4 + ) −  (4) 5 − (4 + ) − 1 = lim   →0− − = lim = −1 →0−  (b) 0 64. (a) − (4) = lim →0− and 1 −1  (4 + ) −  (4) 5 − (4 + ) 0 + = lim (4) = lim   →0+ →0+ 1 − (1 − ) 1 = lim =1 = lim →0+ (1 − ) →0+ 1 −   0 if  ≤ 0   5− if 0    4 (c)  () =   1(5 − ) if  ≥ 4 At 4 we have lim () = lim (5 − ) = 1 and lim () = lim →4− →4− →4+ →4+ 1 = 1, so lim  () = 1 =  (4) and  is →4 5− continuous at 4. Since  (5) is not defined,  is discontinuous at 5. These expressions show that  is continuous on the intervals (−∞ 0), (0 4), (4 5) and (5 ∞). Since lim  () = lim (5 − ) = 5 6= 0 = lim  (), lim  () does →0+ →0+ →0− →0 not exist, so  is discontinuous (and therefore not differentiable) at 0. 0 0 (4) 6= + (4), and from (c),  is not differentiable at 0 or 5. (d) From (a),  is not differentiable at 4 since − 65. These graphs are idealizations conveying the spirit of the problem. In reality, changes in speed are not instantaneous, so the graph in (a) would not have corners and the graph in (b) would be continuous. (a) (b) 66. (a) c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° CHAPTER 2 REVIEW (b) The initial temperature of the water is close to room temperature because of ¤ 151 (c) the water that was in the pipes. When the water from the hot water tank starts coming out,   is large and positive as  increases to the temperature of the water in the tank. In the next phase,   = 0 as the water comes out at a constant, high temperature. After some time,   becomes small and negative as the contents of the hot water tank are exhausted. Finally, when the hot water has run out,   is once again 0 as the water maintains its (cold) temperature. In the right triangle in the diagram, let ∆ be the side opposite angle  and ∆ 67. the side adjacent to angle . Then the slope of the tangent line  is  = ∆∆ = tan . Note that 0    2 . We know (see Exercise 19) that the derivative of () = 2 is  0 () = 2. So the slope of the tangent to the curve at the point (1 1) is 2. Thus,  is the angle between 0 and 2 whose tangent is 2; that is,  = tan−1 2 ≈ 63◦ . 2 Review 1. False. Limit Law 2 applies only if the individual limits exist (these don’t). 2. False. Limit Law 5 cannot be applied if the limit of the denominator is 0 (it is). 3. True. Limit Law 5 applies. 4. False. 2 − 9 is not defined when  = 3, but  + 3 is. −3 5. True. 6. True. lim →3 2 − 9 ( + 3)( − 3) = lim = lim ( + 3) →3 →3 −3 ( − 3) The limit doesn’t exist since  ()() doesn’t approach any real number as  approaches 5. (The denominator approaches 0 and the numerator doesn’t.) 7. False. Consider lim →5 ( − 5) sin( − 5) or lim . The first limit exists and is equal to 5. By Example 2.2.3, we know that →5 −5 −5 the latter limit exists (and it is equal to 1). 8. False. If  () = 1, () = −1, and  = 0, then lim () does not exist, lim () does not exist, but →0 →0 lim [() + ()] = lim 0 = 0 exists. →0 9. True. →0 Suppose that lim [ () + ()] exists. Now lim () exists and lim () does not exist, but → → → lim () = lim {[ () + ()] −  ()} = lim [ () + ()] − lim () [by Limit Law 2], which exists, and → → → → we have a contradiction. Thus, lim [ () + ()] does not exist. → c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 152 ¤ 10. False. CHAPTER 2 LIMITS AND DERIVATIVES  Consider lim [()()] = lim ( − 6) →6 →6 so  (6)(6) 6= 1. 11. True. 12. False.  1 . It exists (its value is 1) but  (6) = 0 and (6) does not exist, −6 A polynomial is continuous everywhere, so lim () exists and is equal to (). → Consider lim [() − ()] = lim →0 →0 approaches ∞.   1 1 . This limit is −∞ (not 0), but each of the individual functions − 2 4 13. True. See Figure 2.6.8. 14. False. Consider  () = sin  for  ≥ 0. lim  () 6= ±∞ and  has no horizontal asymptote. →∞  1( − 1) if  6= 1 15. False. Consider  () = 16. False. The function  must be continuous in order to use the Intermediate Value Theorem. For example, let  1 if 0 ≤   3  () = There is no number  ∈ [0 3] with  () = 0. −1 if  = 3 17. True. Use Theorem 2.5.8 with  = 2,  = 5, and () = 42 − 11. Note that  (4) = 3 is not needed. 18. True. Use the Intermediate Value Theorem with  = −1,  = 1, and  = , since 3    4. if  = 1 2 19. True, by the definition of a limit with  = 1. 20. False. For example, let  () = 2  + 1 if  6= 0 if  = 0 2   Then ()  1 for all , but lim  () = lim 2 + 1 = 1. →0 21. False. 22. True. 23. False. 24. True. →0 See the note after Theorem 2.8.4.  0() exists ⇒  is differentiable at   2 is the second derivative while 2  2  2  = 0, but = 1. then 2     ⇒  is continuous at  2 ⇒ lim  () =  (). → is the first derivative squared. For example, if  = ,  () = 10 − 102 + 5 is continuous on the interval [0 2],  (0) = 5,  (1) = −4, and  (2) = 989. Since −4  0  5, there is a number  in (0 1) such that  () = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation 10 − 102 + 5 = 0 in the interval (0 1). Similarly, there is a root in (1 2). 25. True. See Exercise 2.5.72(b). 26. False See Exercise 2.5.72(b). c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° CHAPTER 2 REVIEW 1. (a) (i) lim  () = 3 (ii) →2+ ¤ lim  () = 0 →−3+ (iii) lim  () does not exist since the left and right limits are not equal. (The left limit is −2.) →−3 (iv) lim  () = 2 →4 (v) lim  () = ∞ (vi) lim  () = −∞ (vii) lim  () = 4 (viii) lim  () = −1 →0 →2− →∞ →−∞ (b) The equations of the horizontal asymptotes are  = −1 and  = 4. (c) The equations of the vertical asymptotes are  = 0 and  = 2. (d)  is discontinuous at  = −3, 0, 2, and 4. The discontinuities are jump, infinite, infinite, and removable, respectively. 2. →−∞ lim () = −2, →∞ lim  () = −∞, →3+ →3− lim  () = 0, lim  () = ∞, →−3 lim  () = 2,  is continuous from the right at 3 3 3. Since the exponential function is continuous, lim  − = 1−1 = 0 = 1. →1 4. Since rational functions are continuous, lim 2 − 9 →3 2 + 2 − 3 = 32 − 9 32 + 2(3) − 3 = 0 = 0. 12 −3 − 3 −6 3 2 − 9 ( + 3)( − 3) −3 = lim = lim = = = →−3 2 + 2 − 3 →−3 ( + 3)( − 1) →−3  − 1 −3 − 1 −4 2 5. lim 2 − 9 2 − 9 2 + + = −∞ since   0 for 1    3. + 2 − 3 → 0 as  → 1 and 2 + 2 − 3 →1+ 2 + 2 − 3 6. lim   3    − 32 + 3 − 1 + 1 ( − 1)3 + 1 3 − 32 + 3 = lim = lim = lim 2 − 3 + 3 = 3 →0 →0 →0 →0    7. lim Another solution: Factor the numerator as a sum of two cubes and then simplify.   [( − 1) + 1] ( − 1)2 − 1( − 1) + 12 ( − 1)3 + 1 ( − 1)3 + 13 = lim = lim lim →0 →0 →0      2 = lim ( − 1) −  + 2 = 1 − 0 + 2 = 3 →0 ( + 2)( − 2) 2+2 4 1 2 − 4 +2 = lim = lim 2 = = = →2 3 − 8 →2 ( − 2)(2 + 2 + 4) →2  + 2 + 4 4+4+4 12 3 8. lim √ √   4 + = ∞ since ( − 9) → 0 as  → 9 and  0 for  6= 9. →9 ( − 9)4 ( − 9)4 9. lim c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 153 154 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 4− 10. lim →4+ |4 − | →4+ −(4 − ) 4 − 1 11. lim 4− = lim →1 3 + 52 − 6 = lim →1 = lim 1 →4+ −1 = −1 2(2) 4 (2 + 1)(2 − 1) (2 + 1)( + 1)( − 1) (2 + 1)( + 1) = lim = lim = = →1 →1 (2 + 5 − 6) ( + 6)( − 1) ( + 6) 1(7) 7 √ √ √  √ +6− +6− +6+ (  + 6 )2 − 2 √  √ = lim · = lim →3 3 − 32 →3 →3 2 ( − 3) 2 ( − 3) +6+ +6+ 12. lim  + 6 − 2 −(2 −  − 6) −( − 3)( + 2) √  = lim √  = lim √  →3 2 ( − 3) →3 2 ( − 3) →3 2 ( − 3) +6+ +6+ +6+ = lim = lim →3 2 13. Since  is positive, 5 −( + 2) 5 √  =− =− 9(3 + 3) 54 +6+ √ 2 = || = . Thus,  √ √ √ √ 1 − 92 1 2 − 9 2 − 9 2 1−0 = lim = lim = = lim →∞ 2 − 6 →∞ (2 − 6) →∞ 2 − 6 2−0 2 14. Since  is negative, √ 2 = || = −. Thus,  √ √ √ √ 1 − 92 1 2 − 9 2 − 9 2 1−0 = lim = lim = =− →−∞ 2 − 6 →−∞ (2 − 6)(−) →−∞ −2 + 6 −2 + 0 2 lim 15. Let  = sin . Then as  → − , sin  → 0+ , so  → 0+ . Thus, lim ln(sin ) = lim ln  = −∞. → − 16. lim →−∞ 17. lim →∞ →0+ 0−0−1 −1 1 1 − 22 − 4 (1 − 22 − 4 )4 14 − 22 − 1 = = = = lim = lim 4 4 4 →−∞ →−∞ 5 +  − 3 (5 +  − 3 ) 54 + 13 − 3 0+0−3 −3 3 √ √ 2   + 4 + 1 −  2 + 4 + 1 +  (2 + 4 + 1) − 2 ·√ = lim √ →∞ →∞ 1 2 + 4 + 1 +  2 + 4 + 1 +    √ (4 + 1) divide by  = 2 for   0 = lim √ →∞ ( 2 + 4 + 1 + ) √  2 + 4 + 1 −  = lim 4 + 1 4 4+0 = =2 = lim  = √ →∞ 2 1+0+0+1 1 + 4 + 12 + 1 2 18. Let  =  − 2 = (1 − ). Then as  → ∞,  → −∞, and lim − = lim  = 0. →∞ →−∞ 19. Let  = 1. Then as  → 0+ ,  → ∞ , and lim tan−1 (1) = lim tan−1  = →0+ 20. lim →1  1 1 + 2 −1  − 3 + 2  →∞  . 2     1 1 1 −2 = lim + = lim + →1  − 1 →1 ( − 1)( − 2) ( − 1)( − 2) ( − 1)( − 2)   1 1 −1 = lim = = −1 = lim →1 ( − 1)( − 2) →1  − 2 1−2 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° CHAPTER 2 REVIEW  ¤  21. From the graph of  = cos2  2 , it appears that  = 0 is the horizontal asymptote and  = 0 is the vertical asymptote. Now 0 ≤ (cos )2 ≤ 1 ⇒ cos2  1 0 ≤ ≤ 2 2  2  ⇒ 0≤ cos2  1 ≤ 2 . But lim 0 = 0 and →±∞ 2  1 cos2  = 0, so by the Squeeze Theorem, lim = 0. 2 →±∞  →±∞ 2 lim cos2  = ∞ because cos2  → 1 and 2 → 0+ as  → 0, so  = 0 is the →0 2 Thus,  = 0 is the horizontal asymptote. lim vertical asymptote. 22. From the graph of  =  () = √ √ 2 +  + 1 − 2 − , it appears that there are 2 horizontal asymptotes and possibly 2 vertical asymptotes. To obtain a different form for  , let’s multiply and divide it by its conjugate. √ √ √ √  2 +  + 1 + 2 −  (2 +  + 1) − (2 − ) 2 2 √ √ = √  ++1−  − √ 1 () = 2 +  + 1 + 2 −  2 +  + 1 + 2 −  2 + 1 √ = √ 2 +  + 1 + 2 −  Now 2 + 1 √ lim 1 () = lim √ →∞ 2 +  + 1 + 2 −  →∞ 2 + (1)  = lim  →∞ 1 + (1) + (12 ) + 1 − (1) = [since √ 2 =  for   0] 2 = 1, 1+1 so  = 1 is a horizontal asymptote. For   0, we have √ 2 = || = −, so when we divide the denominator by , with   0, we get    √ √ √ √ 1 1 1 2 +  + 1 + 2 −  2 +  + 1 + 2 −  √ =− =− 1+ + 2 + 1−     2 Therefore, 2 + 1 2 + (1)   √ lim 1 () = lim √ = lim  2 2 →−∞ →∞  ++1+  − − 1 + (1) + (12 ) + 1 − (1) →−∞ = 2 = −1 −(1 + 1) so  = −1 is a horizontal asymptote. The domain of  is (−∞ 0] ∪ [1 ∞). As  → 0− ,  () → 1, so √  = 0 is not a vertical asymptote. As  → 1+ ,  () → 3, so  = 1 is not a vertical asymptote and hence there are no vertical asymptotes. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 155 156 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 23. Since 2 − 1 ≤ () ≤ 2 for 0    3 and lim (2 − 1) = 1 = lim 2 , we have lim  () = 1 by the Squeeze Theorem. →1 →1    →1   24. Let () = −2 , () = 2 cos 12 and () = 2 . Then since cos 12  ≤ 1 for  6= 0, we have  () ≤ () ≤ () for  6= 0, and so lim  () = lim () = 0 ⇒ →0 →0 lim () = 0 by the Squeeze Theorem. →0 25. Given   0, we need   0 such that if 0  | − 2|  , then |(14 − 5) − 4|  . But |(14 − 5) − 4|   |−5 + 10|   ⇔ ⇔ |−5| | − 2|   ⇔ | − 2|  5. So if we choose  = 5, then 0  | − 2|   ⇒ |(14 − 5) − 4|  . Thus, lim (14 − 5) = 4 by the definition of a limit. →2 √ √ √ 26. Given   0 we must find   0 so that if 0  | − 0|  , then | 3  − 0|  . Now | 3  − 0| = | 3 |   ⇒ √  √ √ 3 ⇒ | 3  − 0| = | 3 | = 3 ||  3 = . √ 3 || = | 3 |  3 . So take  = 3 . Then 0  | − 0| = ||  3 √ Therefore, by the definition of a limit, lim 3  = 0. →0   27. Given   0, we need   0 so that if 0  | − 2|  , then 2 − 3 − (−2)  . First, note that if | − 2|  1, then −1   − 2  1, so 0   − 1  2 ⇒ | − 1|  2. Now let  = min {2 1}. Then 0  | − 2|    2   − 3 − (−2) = |( − 2)( − 1)| = | − 2| | − 1|  (2)(2) = . ⇒ Thus, lim (2 − 3) = −2 by the definition of a limit. →2 √ 28. Given   0, we need   0 such that if 0   − 4  , then 2  − 4  . This is true  − 4  4 2 . So if we choose  = 4 2 , then 0   − 4    √  lim 2  − 4 = ∞. ⇔ √  − 4  2 ⇔ √ ⇒ 2  − 4  . So by the definition of a limit, →4+ 29. (a)  () = √ − if   0,  () = 3 −  if 0 ≤   3,  () = ( − 3)2 if   3. √ − = 0 (i) lim  () = lim (3 − ) = 3 (ii) lim  () = lim (iii) Because of (i) and (ii), lim  () does not exist. (iv) lim  () = lim (3 − ) = 0 →0+ →0− →0+ →0 →3− 2 (v) lim  () = lim ( − 3) = 0 →3+ →0− →3− (vi) Because of (iv) and (v), lim  () = 0. →3 →3+ (b)  is discontinuous at 0 since lim  () does not exist. →0 (c)  is discontinuous at 3 since  (3) does not exist. 30. (a) () = 2 − 2 if 0 ≤  ≤ 2, () = 2 −  if 2   ≤ 3, () =  − 4 if 3    4, () =  if  ≥ 4. Therefore, lim () = lim →2− →2−   2 − 2 = 0 and lim () = lim (2 − ) = 0. Thus, lim () = 0 =  (2), →2+ →2 →2+ so  is continuous at 2. lim () = lim (2 − ) = −1 and lim () = lim ( − 4) = −1. Thus, →3− →3− →3+ →3+ c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° CHAPTER 2 REVIEW ¤ 157 (b) lim () = −1 = (3), so  is continuous at 3. →3 lim () = lim ( − 4) = 0 and lim () = lim  = . →4− →4− →4+ →4+ Thus, lim () does not exist, so  is discontinuous at 4. But →4 lim () =  = (4), so  is continuous from the right at 4. →4+ 31. sin  and  are continuous on R by Theorem 2.5.7. Since  is continuous on R, sin  is continuous on R by Theorem 2.5.9. Lastly,  is continuous on R since it’s a polynomial and the product sin  is continuous on its domain R by Theorem 2.5.4. 32. 2 − 9 is continuous on R since it is a polynomial and √  is continuous on [0 ∞) by Theorem 2.5.7, so the composition √   2 − 9 is continuous on  | 2 − 9 ≥ 0 = (−∞ −3] ∪ [3 ∞) by Theorem 2.5.9. Note that 2 − 2 6= 0 on this set and √ 2 − 9 so the quotient function () = 2 is continuous on its domain, (−∞ −3] ∪ [3 ∞) by Theorem 2.5.4.  −2 33.  () = 5 − 3 + 3 − 5 is continuous on the interval [1 2],  (1) = −2, and (2) = 25. Since −2  0  25, there is a number  in (1 2) such that  () = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation 5 − 3 + 3 − 5 = 0 in the interval (1 2). 34.  () = cos √  −  + 2 is continuous on the interval [0 1], (0) = 2, and  (1) ≈ −02. Since −02  0  2, there is a number  in (0 1) such that  () = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation √ √ cos  −  + 2 = 0, or cos  =  − 2, in the interval (0 1). 35. (a) The slope of the tangent line at (2 1) is lim →2  () −  (2) 9 − 22 − 1 8 − 22 −2(2 − 4) −2( − 2)( + 2) = lim = lim = lim = lim →2 →2 →2 →2 −2 −2 −2 −2 −2 = lim [−2( + 2)] = −2 · 4 = −8 →2 (b) An equation of this tangent line is  − 1 = −8( − 2) or  = −8 + 17. 36. For a general point with -coordinate , we have  = lim → 2(1 − 3) − 2(1 − 3) 2(1 − 3) − 2(1 − 3) 6( − ) = lim = lim → (1 − 3)(1 − 3)( − ) → (1 − 3)(1 − 3)( − ) − 6 = lim → (1 − 3)(1 − 3) = 6 (1 − 3)2 For  = 0,  = 6 and  (0) = 2, so an equation of the tangent line is  − 2 = 6( − 0) or  = 6 + 2 For  = −1,  = 38 and  (−1) = 12 , so an equation of the tangent line is  − 12 = 38 ( + 1) or  = 38  + 78 . 37. (a)  = () = 1 + 2 + 2 4. The average velocity over the time interval [1 1 + ] is ave =  1 + 2(1 + ) + (1 + )2 4 − 134 10 + 2 10 +  (1 + ) − (1) = = = (1 + ) − 1  4 4 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° [continued] 158 ¤ CHAPTER 2 LIMITS AND DERIVATIVES So for the following intervals the average velocities are: (i) [1 3]:  = 2, ave = (10 + 2)4 = 3 ms (ii) [1 2]:  = 1, ave = (10 + 1)4 = 275 ms (iii) [1 15]:  = 05, ave = (10 + 05)4 = 2625 ms (iv) [1 11]:  = 01, ave = (10 + 01)4 = 2525 ms (b) When  = 1, the instantaneous velocity is lim →0 10 (1 + ) − (1) 10 +  = lim = = 25 ms. →0  4 4 38. (a) When  increases from 200 in3 to 250 in3 , we have ∆ = 250 − 200 = 50 in3 , and since  = 800 , ∆ =  (250) −  (200) = is 800 800 − = 32 − 4 = −08 lbin2 . So the average rate of change 250 200 lbin2 ∆ −08 = = −0016 . ∆ 50 in3 (b) Since  = 800 , the instantaneous rate of change of  with respect to  is lim ∆ →0 ∆ = lim →0  ( + ) −  ( ) 800( + ) − 800 800 [ − ( + )] = lim = lim →0 →0   ( + ) = lim −800 →0 ( + ) =− 800 2 which is inversely proportional to the square of  . () −  (2) 3 − 2 − 4 = lim →2 →2 −2 −2 39. (a)  0 (2) = lim (c) ( − 2)(2 + 2 + 2) = lim (2 + 2 + 2) = 10 →2 →2 −2 = lim (b)  − 4 = 10( − 2) or  = 10 − 16 40. 26 = 64, so  () = 6 and  = 2. 41. (a)  0 () is the rate at which the total cost changes with respect to the interest rate. Its units are dollars(percent per year). (b) The total cost of paying off the loan is increasing by $1200(percent per year) as the interest rate reaches 10%. So if the interest rate goes up from 10% to 11%, the cost goes up approximately $1200. (c) As  increases,  increases. So  0 () will always be positive. 42. 43. 44. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° CHAPTER 2 REVIEW ¤ 159   √ √ 3 − 5( + ) − 3 − 5 3 − 5( + ) + 3 − 5 ( + ) −  ()  = lim 45. (a)  0 () = lim √ →0 →0   3 − 5( + ) + 3 − 5 = lim →0 −5 [3 − 5( + )] − (3 − 5) −5   = lim  = √ √ √ →0 2 3 − 5 3 − 5( + ) + 3 − 5  3 − 5( + ) + 3 − 5 (b) Domain of  : (the radicand must be nonnegative) 3 − 5 ≥ 0 ⇒   5 ≤ 3 ⇒  ∈ −∞ 35 Domain of  0 : exclude 35 because it makes the denominator zero;    ∈ −∞ 35 (c) Our answer to part (a) is reasonable because  0 () is always negative and  is always decreasing. 46. (a) As  → ±∞,  () = (4 − )(3 + ) → −1, so there is a horizontal asymptote at  = −1. As  → −3+ ,  () → ∞, and as  → −3− ,  () → −∞. Thus, there is a vertical asymptote at  = −3. (b) Note that  is decreasing on (−∞ −3) and (−3 ∞), so  0 is negative on those intervals. As  → ±∞,  0 → 0. As  → −3− and as  → −3+ ,  0 → −∞. 4− 4 − ( + ) − ( + ) −  () 3 + ( + ) 3+ (3 + ) [4 − ( + )] − (4 − ) [3 + ( + )] = lim = lim (c)  0 () = lim →0 →0 →0    [3 + ( + )] (3 + ) (12 − 3 − 3 + 4 − 2 − ) − (12 + 4 + 4 − 3 − 2 − ) →0 [3 + ( + )](3 + ) = lim = lim −7 →0  [3 + ( + )] (3 + ) = lim −7 →0 [3 + ( + )] (3 + ) =− 7 (3 + )2 (d) The graphing device confirms our graph in part (b). 47.  is not differentiable: at  = −4 because  is not continuous, at  = −1 because  has a corner, at  = 2 because  is not continuous, and at  = 5 because  has a vertical tangent. 48. The graph of  has tangent lines with positive slope for   0 and negative slope for   0, and the values of  fit this pattern, so  must be the graph of the derivative of the function for . The graph of  has horizontal tangent lines to the left and right of the -axis and  has zeros at these points. Hence,  is the graph of the derivative of the function for . Therefore,  is the graph of  ,  is the graph of  0 , and  is the graph of  00 . c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 160 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 49. Domain: (−∞ 0) ∪ (0 ∞); lim () = 1; lim  () = 0; →0− →0+  0 ()  0 for all  in the domain; lim  0 () = 0; lim  0 () = 1 →−∞ →∞ 50. (a)  0 () is the rate at which the percentage of Americans under the age of 18 is changing with respect to time. Its units are percent per year (%yr). (b) To find  0 (), we use lim →0 For 1950:  0 (1950) ≈  ( + ) −  ()  ( + ) −  () ≈ for small values of .   357 − 311  (1960) −  (1950) = = 046 1960 − 1950 10 For 1960: We estimate  0 (1960) by using  = −10 and  = 10, and then average the two results to obtain a final estimate.  = −10 ⇒  0 (1960) ≈  = 10 ⇒  0 (1960) ≈ 311 − 357  (1950) −  (1960) = = 046 1950 − 1960 −10 340 − 357  (1970) −  (1960) = = −017 1970 − 1960 10 So we estimate that  0 (1960) ≈ 12 [046 + (−017)] = 0145.  1950 1960 1970 1980 1990 2000 2010 0 0460 0145 −0385 −0415 −0115 −0085 −0170  () (c) (d) We could get more accurate values for  0 () by obtaining data for the mid-decade years 1955, 1965, 1975, 1985, 1995, and 2005. 51.  0 () is the rate at which the number of US $20 bills in circulation is changing with respect to time. Its units are billions of bills per year. We use a symmetric difference quotient to estimate  0 (2000).  0 (2000) ≈ 577 − 421 (2005) − (1995) = = 0156 billions of bills per year (or 156 million bills per year). 2005 − 1995 10 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° CHAPTER 2 REVIEW ¤ 0 −16 52. (a) Drawing slope triangles, we obtain the following estimates:  0 (1950) ≈ 11 10 = 011,  (1965) ≈ 10 = −016, and  0 (1987) ≈ 02 = 002. 10 (b) The rate of change of the average number of children born to each woman was increasing by 011 in 1950, decreasing by 016 in 1965, and increasing by 002 in 1987. (c) There are many possible reasons: • In the baby-boom era (post-WWII), there was optimism about the economy and family size was rising. • In the baby-bust era, there was less economic optimism, and it was considered less socially responsible to have a large family. • In the baby-boomlet era, there was increased economic optimism and a return to more conservative attitudes. 53. | ()| ≤ () ⇔ −() ≤  () ≤ () and lim () = 0 = lim −(). → → Thus, by the Squeeze Theorem, lim  () = 0. → 54. (a) Note that  is an even function since () =  (−). Now for any integer , [[]] + [[−]] =  −  = 0, and for any real number  which is not an integer, [[]] + [[−]] = [[]] + (− [[]] − 1) = −1. So lim  () exists (and is equal to −1) → for all values of . (b)  is discontinuous at all integers. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 161 162 ¤ CHAPTER 2 LIMITS AND DERIVATIVES c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. °