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Shop What's New Sign inSolution Manual for Single Variable Calculus: Concepts and Contexts, 4th Edition
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NOT FOR SALE
2
LIMITS AND DERIVATIVES
2.1 The Tangent and Velocity Problems
1. (a) Using S (15> 250), we construct the following table:
w
T
slope = pS T
5
(5> 694)
694250
= 444
= 44=4
515
10
10
(10> 444)
444250
= 194
= 38=8
1015
5
20
(20> 111)
111250
= 139
= 27=8
2015
5
25
(25> 28)
28250
222
2515 = 10 = 22=2
30
(30> 0)
0250
= 250
= 16=6
3015
15
(b) Using the values of w that correspond to the points
closest to S (w = 10 and w = 20), we have
38=8 + (27=8)
= 33=3
2
(c) From the graph, we can estimate the slope of the
tangent line at S to be 300
= 33=3.
9
2530
2. (a) Slope = 2948
= 418
42 36
6
2661
(b) Slope = 2948
= 287
= 71=75
42 38
4
69=67
2948
= 132
= 66
(d) Slope = 3080
44 42
2
2806
= 142
= 71
(c) Slope = 2948
42 40
2
From the data, we see that the patientโs heart rate is decreasing from 71 to 66 heartbeats@minute after 42 minutes.
After being stable for a while, the patientโs heart rate is dropping.
3. (a) | =
{
, S (1> 12 )
1+{
(b) The slope appears to be 14 .
{
T
pS T
(i)
0=5
(0=5> 0=333333)
0=333333
(ii)
0=9
(0=9> 0=473684)
0=263158
(iii)
0=99
(0=99> 0=497487)
0=251256
(iv)
0=999
(0=999> 0=499750)
0=250125
(v)
1=5
(1=5> 0=6)
0=2
(vi)
1=1
(1=1> 0=523810)
0=238095
(vii)
1=01
(1=01> 0=502488)
0=248756
(viii)
1=001
(1=001> 0=500250)
0=249875
(c) | 12 = 14 ({ 1) or | = 14 { + 14 .
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NOT FOR SALE
CHAPTER 2
LIMITS AND
D DERIVATIVES
VES
4. (a) | = cos {, S (0=5> 0)
{
(b) The slope appears to be .
T
pS T
(i)
0
(0> 1)
2
(ii)
0=4
(0=4> 0=309017)
3=090170
(iii)
0=49
(0=49> 0=031411)
3=141076
(iv)
0=499
(0=499> 0=003142)
3=141587
(v)
1
(1> 1)
2
(vi)
0=6
(0=6> 0=309017)
3=090170
(vii)
0=51
(0=51> 0=031411)
3=141076
(viii)
0=501
(0=501> 0=003142)
3=141587
(c) | 0 = ({ 0=5) or | = { + 12 .
(d)
5. (a) | = |(w) = 40w 16w2 . At w = 2, | = 40(2) 16(2)2 = 16. The average velocity between times 2 and 2 + k is
40(2 + k) 16(2 + k)2 16
24k 16k2
|(2 + k) |(2)
yave =
=
=
= 24 16k, if k 6= 0.
(2 + k) 2
k
k
(i) [2> 2=5]: k = 0=5, yave = 32 ft@s
(ii) [2> 2=1]: k = 0=1, yave = 25=6 ft@s
(iii) [2> 2=05]: k = 0=05, yave = 24=8 ft@s
(iv) [2> 2=01]: k = 0=01, yave = 24=16 ft@s
(b) The instantaneous velocity when w = 2 (k approaches 0) is 24 ft@s.
6. (a) | = |(w) = 10w 1=86w2 . At w = 1, | = 10(1) 1=86(1)2 = 8=14. The average velocity between times 1 and 1 + k is
10(1 + k) 1=86(1 + k)2 8=14
6=28k 1=86k2
|(1 + k) |(1)
=
=
= 6=28 1=86k, if k 6= 0.
yave =
(1 + k) 1
k
k
(i) [1> 2]: k = 1, yave = 4=42 m@s
(ii) [1> 1=5]: k = 0=5, yave = 5=35 m@s
(iii) [1> 1=1]: k = 0=1, yave = 6=094 m@s
(iv) [1> 1=01]: k = 0=01, yave = 6=2614 m@s
(v) [1> 1=001]: k = 0=001, yave = 6=27814 m@s
(b) The instantaneous velocity when w = 1 (k approaches 0) is 6=28 m@s.
7. (a) (i) On the interval [1> 3], yave =
v(3) v(1)
10=7 1=4
9=3
=
=
= 4=65 m@s.
31
2
2
(ii) On the interval [2> 3], yave =
v(3) v(2)
10=7 5=1
=
= 5=6 m@s.
32
1
(iii) On the interval [3> 5], yave =
25=8 10=7
15=1
v(5) v(3)
=
=
= 7=55 m@s.
53
2
2
(iv) On the interval [3> 4], yave =
17=7 10=7
v(4) v(3)
=
= 7 m@s.
43
1
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SECTION
TIO 22.11
(b)
THE TANGENT AND
AN VELOCITY PROBLEMS
ยค
Using the points (2> 4) and (5> 23) from the approximate tangent
line, the instantaneous velocity at w = 3 is about
8. (a) (i) v = v(w) = 2 sin w + 3 cos w. On the interval [1> 2], yave =
(ii) On the interval [1> 1=1], yave =
v(1=1) v(1)
1=1 1
(iii) On the interval [1> 1=01], yave =
6=3 m@s.
v(2) v(1)
3 (3)
=
= 6 cm@s.
21
1
3=471 (3)
= 4=71 cm@s.
0=1
v(1=01) v(1)
1=01 1
(iv) On the interval [1> 1=001], yave =
23 4
52
3=0613 (3)
= 6=13 cm@s.
0=01
v(1=001) v(1)
1=001 1
3=00627 (3)
= 6=27 cm@s.
1=001 1
(b) The instantaneous velocity of the particle when w = 1 appears to be about 6=3 cm@s.
9. (a) For the curve | = sin(10@{) and the point S (1> 0):
{
T
pS T
{
T
pS T
2
(2> 0)
0
0=5
(0=5> 0)
0
1=5
(1=5> 0=8660)
1=7321
0=6
(0=6> 0=8660)
2=1651
1=4
(1=4> 0=4339)
1=0847
0=7
(0=7> 0=7818)
2=6061
1=3
(1=3> 0=8230)
2=7433
0=8
(0=8> 1)
5
1=2
(1=2> 0=8660)
4=3301
0=9
(0=9> 0=3420)
1=1
(1=1> 0=2817)
2=8173
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 S 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 T is (1=001> 0=0314) and pS T
31=3794. If { = 0=999, then T is
(0=999> 0=0314) and pS T = 31=4422. The average of these slopes is 31=4108. So we estimate that the slope of the
tangent line at S is about 31=4.
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CHAPTER 2
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LIMITS AND
D DERIVATIVES
VES
2.2 The Limit of a Function
1. As { approaches 2, i ({) approaches 5. [Or, the values of i ({) can be made as close to 5 as we like by taking { sufciently
close to 2 (but { 6= 2).] Yes, the graph could have a hole at (2> 5) and be dened such that i (2) = 3.
2. As { approaches 1 from the left, i ({) approaches 3; and as { approaches 1 from the right, i({) approaches 7. No, the limit
does not exist because the left- and right-hand limits are different.
3. (a) i ({) approaches 2 as { approaches 1 from the left, so lim i ({) = 2.
{1
(b) i ({) approaches 3 as { approaches 1 from the right, so lim i ({) = 3.
{1+
(c) lim i ({) does not exist because the limits in part (a) and part (b) are not equal.
{1
(d) i ({) approaches 4 as { approaches 5 from the left and from the right, so lim i ({) = 4.
{5
(e) i(5) is not dened, so it doesnโt exist.
4. (a) lim i ({) = 3
(b) lim i ({) = 4
{0
(c) lim i ({) = 2
{3
{3+
(d) lim i ({) does not exist because the limits in part (b) and part (c) are not equal.
{3
(e) i (3) = 3
5. (a) lim j(w) = 1
(b) lim j(w) = 2
w0
w0+
(c) lim j(w) does not exist because the limits in part (a) and part (b) are not equal.
w0
(d) lim j(w) = 2
(e) lim j(w) = 0
w2
w2+
(f ) lim j(w) does not exist because the limits in part (d) and part (e) are not equal.
w2
(h) lim j(w) = 3
(g) j(2) = 1
w4
6. (a) k({) approaches 4 as { approaches 3 from the left, so
lim k({) = 4.
{3
(b) k({) approaches 4 as { approaches 3 from the right, so lim k({) = 4.
{3+
(c) lim k({) = 4 because the limits in part (a) and part (b) are equal.
{3
(d) k(3) is not dened, so it doesnโt exist.
(e) k({) approaches 1 as { approaches 0 from the left, so lim k({) = 1.
{0
(f ) k({) approaches 1 as { approaches 0 from the right, so lim k({) = 1.
{0+
(g) lim k({) does not exist because the limits in part (e) and part (f ) are not equal.
{0
(h) k(0) = 1 since the point (0> 1) is on the graph of k.
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SECTION 22.22 TH
THE LIMIT OF A FUNCTION
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(i) Since lim k({) = 2 and lim k({) = 2, we have lim k({) = 2.
{2
{2+
{2
(j) k(2) is not dened, so it doesnโt exist.
(k) k({) approaches 3 as { approaches 5 from the right, so lim k({) = 3.
{5+
(l) k({) does not approach any one number as { approaches 5 from the left, so lim k({) does not exist.
{5
7. From the graph of
;
1 + { if { ? 1
A
?
if 1 { ? 1 ,
i ({) = {2
A
=
2 { if { 1
we see that lim i({) exists for all d except d = 1. Notice that the
{d
right and left limits are different at d = 1.
8. From the graph of
;
1 + sin { if { ? 0
A
?
if 0 { ,
i ({) = cos {
A
=
sin {
if { A
we see that lim i({) exists for all d except d = . Notice that the
{d
right and left limits are different at d = .
9. (a) lim i ({) = 1
{0
(b) lim i({) = 0
{0+
(c) lim i ({) does not exist because the limits
{0
in part (a) and part (b) are not equal.
10. (a) lim i ({) = 1
{0
(b) lim i({) = 1
{0+
(c) lim i ({) does not exist because the limits
{0
in part (a) and part (b) are not equal.
11. (a) lim i ({) = 2
{0
(b) lim i({) = 2
{0+
(c) lim i ({) does not exist because the limits
{0
in part (a) and part (b) are not equal.
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12.
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CHAPTER 2
LIMITS AND
D DERIVATIVES
VES
lim i (w) = 150 mg and lim i (w) = 300 mg. These limits show that there is an abrupt change in the amount of drug in
w12
+
w12
the patientโs bloodstream at w = 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.
13. lim i ({) = 1,
{0
lim i({) = 2, i (0) = 1
{0+
14. lim i ({) = 1, lim i ({) = 2, lim i ({) = 2,
{0
{3
{3+
i (0) = 1, i (3) = 1
15. lim i ({) = 4,
lim i ({) = 2, lim i({) = 2,
{3+
{2
{3
i (3) = 3, i(2) = 1
17. For i ({) =
16. lim i ({) = 2, lim i ({) = 0, lim i ({) = 3,
{0
{0+
{4
lim i ({) = 0, i (0) = 2, i (4) = 1
{4+
{2 2{
:
{2 { 2
18. For i ({) =
{2 2{
:
{2 { 2
{
i ({)
{
i ({)
{
i ({)
{
i ({)
2=5
0=714286
1=9
0=655172
0
0
2
2
2=1
0=677419
1=95
0=661017
0=5
1
1=5
3
2=05
0=672131
1=99
0=665552
0=9
9
1=1
11
2=01
0=667774
1=995
0=666110
0=95
19
1=01
101
2=005
2=001
0=667221
0=666778
1=999
0=666556
0=99
99
1=001
1001
0=999
999
{2 2{
= 0=6ฬ = 23 .
{2 {2 { 2
It appears that lim
{2 2{
does not exist since
{1 {2 { 2
It appears that lim
i ({)
as {
1 and i ({)
as {
1+ .
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SECTION 22.22 TH
THE LIMIT OF A FUNCTION
19. For i (w) =
h5w 1
:
w
w
i (w)
20. For i (k) =
w
i (w)
ยค
(2 + k)5 32
:
k
k
i (k)
k
i (k)
0=5
22=364988
0=5
1=835830
0=5
131=312500
0=5
48=812500
0=1
6=487213
0=1
3=934693
0=1
88=410100
0=1
72=390100
0=01
5=127110
0=01
4=877058
0=01
80=804010
0=01
79=203990
0=001
5=012521
0=001
4=987521
0=001
80=080040
0=001
79=920040
0=0001
5=001250
0=0001
4=998750
0=0001
80=008000
0=0001
79=992000
h5w 1
= 5.
w0
w
{+42
21. For i ({) =
:
{
(2 + k)5 32
= 80.
k0
k
It appears that lim
{
93
It appears that lim
22. For i ({) =
tan 3{
:
tan 5{
i ({)
{
i ({)
{
i({)
1
0=236068
1
0=267949
ยฑ0=2
0=439279
0=5
0=242641
0=5
0=258343
0=566236
0=248457
0=1
ยฑ0=1
0=1
0=251582
0=591893
0=249224
0=05
ยฑ0=05
0=05
0=250786
0=01
0=249844
0=01
0=250156
ยฑ0=01
ยฑ0=001
0=599680
0=599997
{+42
= 0=25 = 14 .
{0
{
It appears that lim
It appears that lim
tan 3{
{0 tan 5{
23. For i ({) =
{6 1
:
{10 1
{
i ({)
{
i ({)
{
i ({)
0=5
0=985337
1=5
0=183369
0=5
0=9
0=719397
1=1
0=484119
0=95
0=660186
1=05
0=99
0=612018
0=999
0=601200
24. For i ({) =
= 0=6 = 35 .
9{ 5{
:
{
{
i ({)
1=527864
0=5
0=227761
0=1
0=711120
0=1
0=485984
0=540783
0=05
0=646496
0=05
0=534447
1=01
0=588022
0=01
0=599082
0=01
0=576706
1=001
0=598800
0=001
0=588906
0=001
0=586669
{6 1
= 0=6 = 35 .
{1 {10 1
It appears that lim
9{ 5{
= 0=59. Later we will be able
{0
{
It appears that lim
to show that the exact value is ln(9@5).
25. (a) From the graphs, it seems that lim
{0
cos 2{ cos {
= 1=5.
{2
(b)
{
i ({)
ยฑ0=1
1=493759
ยฑ0=01
1=499938
ยฑ0=001
1=499999
ยฑ0=0001
1=500000
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CHAPTER 2
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VES
26. (a) From the graphs, it seems that lim
sin {
{0 sin {
(b)
= 0=32.
{
i ({)
ยฑ0=1
0=323068
ยฑ0=01
0=318357
ยฑ0=001
ยฑ0=0001
0=318310
0=318310
Later we will be able to show that
the exact value is
1@{
27. (a) Let k({) = (1 + {)
{
0=001
0=0001
0=00001
0=000001
0=000001
0=00001
0=0001
0=001
.
1
.
(b)
k({)
2=71964
2=71842
2=71830
2=71828
2=71828
2=71827
2=71815
2=71692
It appears that lim (1 + {)1@{
{0
2=71828, which is approximately h.
In Section 3.7 we will see that the value of the limit is exactly h.
28. For the curve | = 2{ and the points S (0> 1) and T({> 2{ ):
{
T
pS T
0=1
0=01
0=001
0=0001
(0=1> 1=0717735)
(0=01> 1=0069556)
(0=001> 1=0006934)
(0=0001> 1=0000693)
0=71773
0=69556
0=69339
0=69317
The slope appears to be about 0=693.
29. For i ({) = {2 (2{@1000):
(a)
{
i ({)
1
0=8
0=6
0=4
0=2
0=1
0=998000
0=638259
0=358484
0=158680
0=038851
0=008928
0=05
0=001465
It appears that lim i ({) = 0.
(b)
{
i ({)
0=04
0=02
0=01
0=005
0=003
0=000572
0=000614
0=000907
0=000978
0=000993
0=001
0=001000
It appears that lim i ({) = 0=001.
{0
{0
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SECTION 22.22 TH
THE LIMIT OF A FUNCTION
30. For k({) =
(a)
tan { {
:
{3
{
k({)
1=0
0=5
0=1
0=05
0=01
0=005
0=55740773
0=37041992
0=33467209
0=33366700
0=33334667
0=33333667
(b) It seems that lim k({) = 13 .
{0
(c)
{
k({)
0=001
0=0005
0=0001
0=00005
0=00001
0=000001
0=33333350
0=33333344
0=33333000
0=33333600
0=33300000
0=00000000
ยค
95
Here the values will vary from one
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.
31. We need to have 5=8 ? {3 3{ + 4 ? 6=2. From the graph we obtain the approximate points of intersection S (1=9774> 5=8)
and T(2=0219> 6=2). So if { is within 0=021 of 2, then | will be within 0=2 of 6. If we must have {3 3{ + 4 within 0=1 of 6,
we get S (1=9888> 5=9) and T(2=0110> 6=1). We would then need { to be within 0=011 of 2.
{3 1
.
{1
32. (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=92531
5=99250
5=99925
6=07531
6=00750
6=00075
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CHAPTER 2
LIMITS AND
D DERIVATIVES
VES
{3 1
(b) We need to have 5=5 ?
? 6=5. From the graph we obtain the approximate points of intersection S (0=9314> 5=5)
{1
and T(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 [i ({) + 5j({)] = lim i ({) + lim [5j({)]
[Limit Law 1]
= lim i ({) + 5 lim j({)
[Limit Law 3]
{2
{2
{2
{2
{2
= 4 + 5(2) = 6
(b) lim [j({)]3 =
{2
k
l3
lim j({)
[Limit Law 6]
{2
= ( 2)3 = 8
(c) lim
{2
t
s
i ({) = lim i ({)
[Limit Law 11]
{2
=
4=2
lim [3i ({)]
3i ({)
{2
=
{2 j({)
lim j({)
(d) lim
[Limit Law 5]
{2
3 lim i ({)
=
{2
[Limit Law 3]
lim j({)
{2
=
3(4)
= 6
2
(e) Because the limit of the denominator is 0, we canโt use Limit Law 5. The given limit, lim
j({)
{2 k({)
, does not exist because the
denominator approaches 0 while the numerator approaches a nonzero number.
lim [j({) k({)]
j({) k({)
{2
=
{2
i ({)
lim i ({)
[Limit Law 5]
(f ) lim
{2
lim j({) ยท lim k({)
=
{2
{2
[Limit Law 4]
lim i ({)
{2
=
2 ยท 0
=0
4
2. (a) lim [i({) + j({)] = lim i ({) + lim j({) = 2 + 0 = 2
{2
{2
{2
(b) lim j({) does not exist since its left- and right-hand limits are not equal, so the given limit does not exist.
{1
(c) lim [i({)j({)] = lim i ({) ยท lim j({) = 0 ยท 1=3 = 0
{0
{0
{0
(d) Since lim j({) = 0 and j is in the denominator, but lim i ({) = 1 6= 0, the given limit does not exist.
{1
{1
INSTRUCTOR USE ONLY
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SECTION
N 22.33
CALCULATING LIMITS USING THE LIMIT LAWS
ยค
lk
k
l
(e) lim {3 i ({) = lim {3 lim i ({) = 23 ยท 2 = 16
{2
(f ) lim
{1
{2
{2
t
s
3 + i ({) = 3 + lim i ({) = 3 + 1 = 2
{1
3. lim (3{4 + 2{2 { + 1) = lim 3{4 + lim 2{2 lim { + lim 1
{2
{2
{2
{2
[Limit Laws 1 and 2]
{2
= 3 lim {4 + 2 lim {2 lim { + lim 1
[3]
= 3(2)4 + 2(2)2 (2) + (1)
[9, 8, and 7]
{2
{2
{2
{2
= 48 + 8 + 2 + 1 = 59
4. lim (w2 + 1)3 (w + 3)5 = lim (w2 + 1)3 ยท lim (w + 3)5
w1
w1
=
3
5
lim (w2 + 1) ยท lim (w + 3)
[6]
3
5
lim w + lim 1 ยท lim w + lim 3
[1]
w1
=
[Limit Law 4]
w1
w1
2
w1
w1
w1
w1
3
= (1)2 + 1 ยท [1 + 3]5 = 8 ยท 32 = 256
[9, 7, and 8]
5. lim (1 + 3 { ) (2 6{2 + {3 ) = lim (1 + 3 { ) ยท lim (2 6{2 + {3 )
{8
{8
[Limit Law 4]
{8
= lim 1 + lim 3 { ยท lim 2 6 lim {2 + lim {3
{8
{8
{8
{8
{8
3
2
3
= 1+ 8 ยท 26ยท8 +8
[1, 2, and 3]
[7, 10, 9]
= (3)(130) = 390
6. lim
x2
t
x4 + 3x + 6 =
lim (x4 + 3x + 6)
[11]
x2
=
t
lim x4 + 3 lim x + lim 6
x2
x2
x2
t
(2)4 + 3 (2) + 6
= 16 6 + 6 = 16 = 4
[9, 8, and 7]
=
u
7. lim
{2
2{2 + 1
=
3{ 2
[1, 2, and 3]
u
2{2 + 1
{2 3{ 2
y
x lim (2{2 + 1)
x {2
=w
lim (3{ 2)
[Limit Law 11]
lim
[5]
{2
y
x 2 lim {2 + lim 1
x {2
{2
=w
3 lim { lim 2
v
=
{2
[1, 2, and 3]
{2
2(2)2 + 1
=
3(2) 2
u
9
3
=
4
2
[9, 8, and 7]
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CHAPTER 2
LIMITS AND
D DERIVATIVES
VES
8. (a) The left-hand side of the equation is not dened 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 nding lim i ({), we never consider { = d.
{d
{2 6{ + 5
({ 5)({ 1)
= lim
= lim ({ 1) = 5 1 = 4
{5
{5
{5
{5
{5
9. lim
10. lim
{2 4{
{4 {2 3{ 4
= lim
{({ 4)
{4 ({ 4)({ + 1)
= lim
{
{4 { + 1
{2 5{ + 6
does not exist since { 5
{5
{5
=
4
4
=
4+1
5
0, but {2 5{ + 6
11. lim
6 as {
5.
2{2 + 3{ + 1
(2{ + 1)({ + 1)
2{ + 1
2(1) + 1
1
1
= lim
= lim
=
=
=
{1 {2 2{ 3
{1 ({ 3)({ + 1)
{1 { 3
1 3
4
4
12. lim
13. lim
w2 9
w3 2w2 + 7w + 3
14. lim
{2 4{
{1 {2 3{ 4
= lim
(w + 3)(w 3)
w3 (2w + 1)(w + 3)
= lim
w3
w3 2w + 1
does not exist since {2 3{ 4
=
3 3
6
6
=
=
2(3) + 1
5
5
0 but {2 4{
5 as {
1.
(4 + k)2 16
(16 + 8k + k2 ) 16
8k + k2
k(8 + k)
= lim
= lim
= lim
= lim (8 + k) = 8 + 0 = 8
k0
k0
k0
k0
k0
k
k
k
k
15. lim
8 + 12k + 6k2 + k3 8
(2 + k)3 8
12k + 6k2 + k3
= lim
= lim
16. lim
k0
k0
k0
k
k
k
2
= lim 12 + 6k + k = 12 + 0 + 0 = 12
k0
17. 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
1+k1
1+k1
1+k+1
(1 + k) 1
k
= lim
= lim
ยท
18. lim
= lim
k0
k0
k0
k0
k
k
1+k+1
k 1+k+1
k 1+k+1
1
1
1
=
=
= lim
k0
2
1+k+1
1+1
1
{+4
1
+
{+4
1
1
1
4
{
19. lim
= lim 4{ = lim
= lim
=
=
{4 4 + {
{4 4 + {
{4 4{(4 + {)
{4 4{
4(4)
16
20.
{2 + 2{ + 1
({ + 1)2
({ + 1)2
{+1
0
= lim
= lim
= lim
=
=0
4
2
2
2
{1
{1 ({ + 1)({ 1)
{1 ({ + 1)({ + 1)({ 1)
{1 ({2 + 1)({ 1)
{ 1
2(2)
lim
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SECTION
N 22.33
CALCULATING LIMITS USING THE LIMIT LAWS
ยค
4 {
(4 { )(4 + { )
16 {
21. lim
= lim
= lim
{16 16{ {2
{16 (16{ {2 )(4 +
{ ) {16 {(16 {)(4 + { )
= lim
1
{16 {(4 +
22. lim
w0
23. lim
w0
1
1
2
w
w +w
1
1
w
w 1+w
1
1
1
=
=
=
16(8)
128
{)
16 4 + 16
2
w +w w
w2
1
1
= lim
= lim
= lim
=
=1
2
w0 w(w + w)
w0 w ยท w(w + 1)
w0 w + 1
0+1
1 1+w 1+ 1+w
1 1+w
w
= lim
= lim
= lim
w0 w
w0
w0 w
1+w
w w+1 1+ 1+w
1+w 1+ 1+w
1
1
1
=
=
= lim
w0
2
1+w 1+ 1+w
1+0 1+ 1+0
{2 + 9 5
{2 + 9 + 5
{2 + 9 5
({2 + 9) 25
= lim
24. lim
= lim
{4
{4
{4 ({ + 4)
{+4
({ + 4) {2 + 9 + 5
{2 + 9 + 5
{2 16
({ + 4)({ 4)
= lim
2
{4 ({ + 4)
{4
{ +9+5
({ + 4) {2 + 9 + 5
= lim
{4
4
8
4 4
=
=
=
= lim
{4
5+5
5
16 + 9 + 5
{2 + 9 + 5
25. (a)
(b)
{
lim
1 + 3{ 1
{0
(c) lim
{0
2
3
{
i ({)
0=001
0=0001
0=00001
0=000001
0=000001
0=00001
0=0001
0=001
0=6661663
0=6666167
0=6666617
0=6666662
0=6666672
0=6666717
0=6667167
0=6671663
The limit appears to be
2
.
3
{ 1 + 3{ + 1
{ 1 + 3{ + 1
1 + 3{ + 1
{
= lim
ยท
= lim
{0
{0
(1 + 3{) 1
3{
1 + 3{ 1
1 + 3{ + 1
1
lim
1 + 3{ + 1
{0
3
1 t
=
lim (1 + 3{) + lim 1
{0
{0
3
1 t
lim 1 + 3 lim { + 1
=
{0
{0
3
=
1
1+3ยท0+1
3
1
2
= (1 + 1) =
3
3
=
[Limit Law 3]
[1 and 11]
[1, 3, and 7]
[7 and 8]
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CHAPTER 2
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LIMITS AND
ND DERIVATIVES
VE
26. (a)
(b)
3+{ 3
{0
{
lim
0=29
{
i ({)
0=001
0=0001
0=00001
0=000001
0=000001
0=00001
0=0001
0=001
0=2886992
0=2886775
0=2886754
0=2886752
0=2886751
0=2886749
0=2886727
0=2886511
The limit appears to be approximately 0=2887.
(3 + {) 3
3+{ 3
3+{+ 3
1
= lim
= lim
(c) lim
ยท
{0
{0
{0
{
3+{+ 3
{ 3+{+ 3
3+{+ 3
=
lim 1
{0
lim 3 + { + lim 3
{0
= t
[Limit Laws 5 and 1]
{0
1
lim (3 + {) +
{0
3
[7 and 11]
1
=
3+0+ 3
1
=
2 3
[1, 7, and 8]
27. Let i({) = {2 , j({) = {2 cos 20{ and k({) = {2 . Then
1 cos 20{ 1 {2 {2 cos 20{ {2
i ({) j({) k({).
So since lim i ({) = lim k({) = 0, by the Squeeze Theorem we have
{0
{0
lim j({) = 0.
{0
{3 + {2 sin(@{), and k({) = {3 + {2 . Then
1 sin(@{) 1 {3 + {2 {3 + {2 sin(@{) {3 + {2
28. Let i({) = {3 + {2 , j({) =
i ({) j({) k({). So since lim i ({) = lim k({) = 0, by the Squeeze Theorem
{0
{0
we have lim j({) = 0.
{0
29. We have lim (4{ 9) = 4(4) 9 = 7 and lim {2 4{ + 7 = 42 4(4) + 7 = 7. Since 4{ 9 i ({) {2 4{ + 7
{4
{4
for { 0, lim i ({) = 7 by the Squeeze Theorem.
{4
30. We have lim (2{) = 2(1) = 2 and lim ({4 {2 + 2) = 14 12 + 2 = 2. Since 2{ j({) {4 {2 + 2 for all {,
{1
{1
lim j({) = 2 by the Squeeze Theorem.
{1
31. 1 cos(2@{) 1
{4 {4 cos(2@{) {4 . Since lim {4 = 0 and lim {4 = 0, we have
lim {4 cos(2@{) = 0 by the Squeeze Theorem.
{0
{0
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0
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SECTION 22.33
h1 hsin(@{) h1
32. 1 sin(@{) 1
CALCULATING LIMITS U
USING THE LIMIT LAWS
{/h { hsin(@{) { h. Since lim ( {/h) = 0 and
{0+
k
l
{ hsin(@{) = 0 by the Squeeze Theorem.
lim ( { h) = 0, we have lim
{0+
ยค
{0+
+
33. |{ 3| =
{3
if { 3 0
({ 3)
if { 3 ? 0
+
=
{3
if { 3
3{
if { ? 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
+
34. |{ + 6| =
{+6
if { + 6 0
({ + 6)
if { + 6 ? 0
+
=
{+6
if { 6
({ + 6)
if { ? 6
Weโll 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
35. Since |{| = { for { ? 0, we have lim
{0
lim
{6
2{ + 12
2({ + 6)
= lim
= 2
|{ + 6|
{6 ({ + 6)
2{ + 12
does not exist.
|{ + 6|
1
1
{
|{|
= lim
{0
1
1
{
{
= lim
{0
2
, which does not exist since the
{
denominator approaches 0 and the numerator does not.
36. Since |{| = { for { ? 0, we have lim
{2
2 |{|
2 ({)
2+{
= lim
= lim
= lim 1 = 1.
{2
{2 2 + {
{2
2+{
2+{
37. (a) (i) lim j({) = lim { = 1
{1
{1
(ii) lim j({) = lim (2 {2 ) = 2 12 = 1. Since lim j({) = 1 and lim j({) = 1, we have lim j({) = 1.
{1+
{1+
{1
{1+
{1
Note that the fact j(1) = 3 does not affect the value of the limit.
(iii) When { = 1, j({) = 3, so j(1) = 3.
(iv) lim j({) = lim (2 {2 ) = 2 22 = 2 4 = 2
{2
{2
(v) lim j({) = lim ({ 3) = 2 3 = 1
{2+
{2+
(vi) lim j({) does not exist since lim j({) 6= lim j({).
{2
(b)
;
{
A
A
A
A
?3
j({) =
A
2 {2
A
A
A
=
{3
{2
{2+
if { ? 1
if { = 1
if 1 ? { 2
if { A 2
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CHAPTER 2
LIMITS AND
ND DERIVATIVES
VE
{2 1
{2 1
= lim
= lim ({ + 1) = 2
{1+ |{ 1|
{1+ { 1
{1+
38. (a) (i) lim I ({) = lim
{1+
(c)
{2 1
{2 1
= lim
= lim ({ + 1) = 2
{1 |{ 1|
{1 ({ 1)
{1
(ii) lim I ({) = lim
{1
(b) No, lim I ({) does not exist since lim I ({) 6= lim I ({).
{1
{1+
39. (a) (i) [[{]] = 2 for 2 { ? 1, so
(ii) [[{]] = 3 for 3 { ? 2, so
{1
lim [[{]] =
{2+
lim [[{]] =
{2
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) [[{]] = q 1 for q 1 { ? q, so lim [[{]] = lim (q 1) = q 1.
{q
{q
(ii) [[{]] = q for q { ? q + 1, so lim [[{]] = lim q = q.
{q+
{q+
(c) lim [[{]] exists d is not an integer.
{d
40. (a) See the graph of | = cos {.
Since 1 cos { ? 0 on [> @2), we have | = i({) = [[cos {]] = 1
on [> @2).
Since 0 cos { ? 1 on [@2> 0)
on [@2> 0)
(0> @2], we have i({) = 0
(0> @2].
Since 1 cos { ? 0 on (@2> ], we have i ({) = 1 on (@2> ].
Note that i(0) = 1.
(b) (i) lim i ({) = 0 and lim i ({) = 0, so lim i ({) = 0.
{0
{0
{0+
(ii) As {
(@2) , i ({)
0, so
(iii) As {
(@2)+ , i ({)
1, so
lim
{(@2)
lim
i ({) = 0.
{(@2)+
i ({) = 1.
(iv) Since the answers in parts (ii) and (iii) are not equal, lim i ({) does not exist.
{@2
(c) lim i ({) exists for all d in the open interval (> ) except d = @2 and d = @2.
{d
41. The graph of i ({) = [[{]] + [[{]] is the same as the graph of j({) = 1 with holes at each integer, since i (d) = 0 for any
integer d. Thus, lim i ({) = 1 and lim i ({) = 1, so lim i ({) = 1. However,
{2
{2+
{2
i (2) = [[2]] + [[2]] = 2 + (2) = 0, so lim i ({) 6= i (2).
{2
#
42. lim
yf
O0
u
y2
1 2
f
$
= O0 1 1 = 0. As the velocity approaches the speed of light, the length approaches 0.
A left-hand limit is necessary since O is not dened for y A f.
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CALCULATING LIMITS U
USING THE LIMIT LAWS
ยค
43. Since s({) is a polynomial, s({) = d0 + d1 { + d2 {2 + ยท ยท ยท + dq {q . Thus, by the Limit Laws,
lim s({) = lim d0 + d1 { + d2 {2 + ยท ยท ยท + dq {q = d0 + d1 lim { + d2 lim {2 + ยท ยท ยท + dq lim {q
{d
{d
{d
{d
{d
= d0 + d1 d + d2 d2 + ยท ยท ยท + dq dq = s(d)
Thus, for any polynomial s, lim s({) = s(d).
{d
44. Let u({) =
s({)
where s({) and t({) are any polynomials, and suppose that t(d) 6= 0. Thus,
t({)
lim s({)
s({)
{d
=
{d t({)
lim t ({)
lim u({) = lim
{d
[Limit Law 5] =
{d
45. lim [i({) 8] = lim
{1
{1
s(d)
t(d)
[Exercise 43] = u(d).
i({) 8
i ({) 8
ยท ({ 1) = lim
ยท lim ({ 1) = 10 ยท 0 = 0.
{1
{1
{1
{1
Thus, lim i ({) = lim {[i ({) 8] + 8} = lim [i ({) 8] + lim 8 = 0 + 8 = 8.
{1
{1
Note: The value of lim
{1
lim
{1
{1
{1
i ({) 8
does not affect the answer since itโs multiplied by 0. Whatโs important is that
{1
i ({) 8
exists.
{1
46. (a) lim i ({) = lim
{0
{0
i ({)
i ({) 2
ยท
{
= lim 2 ยท lim {2 = 5 ยท 0 = 0
2
{0 {
{0
{
i ({)
i ({)
i ({)
= lim
ยท
{
= lim 2 ยท lim { = 5 ยท 0 = 0
{0 {
{0
{0 {
{0
{2
(b) lim
47. Let i({) = [[{]] and j({) = [[{]]. Then lim i({) and lim j({) do not exist
{3
{3
[Example 9]
but lim [i ({) + j({)] = lim ([[{]] [[{]]) = lim 0 = 0.
{3
{3
{3
48. Let i({) = K({) and j({) = 1 K({), where K is the Heaviside function dened in Example 6 in Section 2.2.
Thus, either i or j is 0 for any value of {. Then lim i ({) and lim j({) do not exist, but lim [i ({)j({)] = lim 0 = 0.
{0
{0
{0
{0
49. Since the denominator approaches 0 as {
0 as {
2, the limit will exist only if the numerator also approaches
2. In order for this to happen, we need lim 3{2 + d{ + d + 3 = 0
{2
3(2)2 + d(2) + d + 3 = 0 12 2d + d + 3 = 0 d = 15. With d = 15, the limit becomes
3({ + 2)({ + 3)
3({ + 3)
3(2 + 3)
3{2 + 15{ + 18
3
= lim
= lim
=
=
= 1.
{2
{2 ({ 1)({ + 2)
{2 { 1
{2 + { 2
2 1
3
lim
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CHAPTER 2
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LIMITS AND
ND DERIVATIVES
VE
50. Solution 1: First, we nd the coordinates of S and T as functions of u. Then we can nd the equation of the line determined
by these two points, and thus nd the {-intercept (the point U), and take the limit as u
0. The coordinates of S are (0> u).
The point T is the point of intersection of the two circles {2 + | 2 = u2 and ({ 1)2 + | 2 = 1. Eliminating | from these
equations, we get u2 {2 = 1 ({ 1)2
u2 = 1 + 2{ 1 { = 12 u2 . Substituting back into the equation of the
shrinking circle to nd the |-coordinate, we get
1 2 2
u
+ | 2 = u2
2
(the positive |-value). So the coordinates of T are
|u =
u
t
1 14 u2 u
1 2
u 0
2
1 2
u >u
2
t
| 2 = u2 1 14 u2
| = u 1 14 u2
t
1 14 u2 . The equation of the line joining S and T is thus
({ 0). We set | = 0 in order to nd the {-intercept, and get
1 2
u
2
{ = u t
=
u
1 14 u2 1
Now we take the limit as u
12 u2
t
1 14 u2 + 1
1 14 u2 1
=2
t
1 14 u2 + 1
t
1 14 u2 + 1 = lim 2 1 + 1 = 4.
0+ : lim { = lim 2
u0+
u0+
u0+
So the limiting position of U is the point (4> 0).
Solution 2: We add a few lines to the diagram, as shown. Note that
S TV = 90 (subtended by diameter S V). So VTU = 90 = RTW
(subtended by diameter RW ). It follows that RTV = W TU. Also
S VT = 90 VS T = RUS . Since 4TRV is isosceles, so is
4TW U, implying that TW = W U. As the circle F2 shrinks, the point T
plainly approaches the origin, so the point U must approach a point twice
as far from the origin as W , that is, the point (4> 0), as above.
2.4 Continuity
1. From Denition 1, lim i ({) = i (4).
{4
2. The graph of i has no hole, jump, or vertical asymptote.
3. (a) i is discontinuous at 4 since i (4) is not dened and at 2, 2, and 4 since the limit does not exist (the left and right
limits are not the same).
(b) i is continuous from the left at 2 since
lim i ({) = i (2). i is continuous from the right at 2 and 4 since
{2
lim i({) = i (2) and lim i ({) = i (4). It is continuous from neither side at 4 since i (4) is undened.
{2+
{4+
4. j is continuous on [4> 2), (2> 2), [2> 4), (4> 6), and (6> 8).
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SECTION 2.4 CONTINUITY
SECT
5. The graph of | = i ({) must have a discontinuity at
105
6. The graph of | = i ({) must have discontinuities
at { = 1 and { = 4. It must show that
{ = 2 and must show that lim i ({) = i (2).
{2+
lim i ({) = i (1) and lim i({) = i(4).
{1
7. The graph of | = i ({) must have a removable
{4+
8. The graph of | = i ({) must have a discontinuity
discontinuity (a hole) at { = 3 and a jump discontinuity
at { = 5.
ยค
at { = 2 with
lim i ({) 6= i (2) and
{2
lim i ({) 6= i(2). It must also show that
{2+
lim i ({) = i (2) and lim i ({) 6= i (2).
{2
{2+
(b) There are discontinuities at times w = 1, 2, 3, and 4. A person
9. (a)
parking in the lot would want to keep in mind that the charge will
jump at the beginning of each hour.
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 specic time changes smoothly as the distance due west from New York City increases,
without any instantaneous jumps.
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LIMITS AND
ND DERIVATIVES
VE
(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 โ 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 denition of current.
11. Since i and j are continuous functions,
lim [2i ({) j({)] = 2 lim i ({) lim j({)
{3
{3
{3
= 2i(3) j(3)
[by Limit Laws 2 and 3]
[by continuity of i and j at { = 3]
= 2 ยท 5 j(3) = 10 j(3)
Since it is given that lim [2i ({) j({)] = 4, we have 10 j(3) = 4, so j(3) = 6.
{3
2
2
lim (2w 3w )
2 lim w 3 lim w
2(1) 3(1)2
1
2w 3w2
w1
w1
w1
12. lim k(w) = lim
=
=
=
=
= k(1).
w1
w1 1 + w3
lim (1 + w3 )
lim 1 + lim w3
1 + (1)3
2
w1
w1
w1
By the denition of continuity, k is continuous at d = 1.
13. lim i ({) = lim
{1
{1
4
{ + 2{3 =
lim { + 2 lim {3
{1
4
{1
4
= 1 + 2(1)3 = (3)4 = 81 = i(1).
By the denition of continuity, i is continuous at d = 1.
14. For d ? 3, we have
lim j({) = lim 2
3{
= 2 lim
3{
{d
{d
{d
=2
=2
=2
t
lim (3 {)
[11]
lim 3 lim {
[2]
{d
t
[Limit Law 3]
{d
3d
{d
[7 and 8]
= j(d)
So j is continuous at { = d for every d in (> 3). Also, lim j({) = 0 = j(3), so j is continuous from the left at 3.
{3
Thus, j is continuous on (> 3]=
15. i ({) =
+ {
h
if { ? 0
{2
if { 0
The left-hand limit of i at d = 0 is lim i ({) = lim h{ = 1. The
{0
{0
right-hand limit of i at d = 0 is lim i ({) = lim {2 = 0. Since these
{0+
{0+
limits are not equal, lim i ({) does not exist and i is discontinuous at 0=
{0
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SECT
; 2
?{ {
{2 1
16. i ({) =
=
1
ยค
107
if { 6= 1
if { = 1
{2 {
{({ 1)
{
1
= lim
= lim
= ,
{1 {2 1
{1 ({ + 1)({ 1)
{1 { + 1
2
lim i ({) = lim
{1
but i (1) = 1, so i is discontinous at 1=
;
cos {
A
?
17. i ({) = 0
A
=
1 {2
if { ? 0
if { = 0
if { A 0
lim i ({) = 1, but i (0) = 0 6= 1, so i is discontinuous at 0.
{0
; 2
? 2{ 5{ 3
{3
18. i ({) =
=
6
if { 6= 3
if { = 3
2{2 5{ 3
(2{ + 1)({ 3)
= lim
= lim (2{ + 1) = 7,
{3
{3
{3
{3
{3
lim i ({) = lim
{3
but i (3) = 6, so i is discontinuous at 3.
19. By Theorem 5, the polynomials {2 and 2{ 1 are continuous on (> ). By Theorem 7, the root function
continuous on [0> ). By Theorem 9, the composite function
By part 1 of Theorem 4, the sum U({) = {2 +
{ is
2{ 1 is continuous on its domain, 12 > .
2{ 1 is continuous on 12 > .
20. By Theorem 7, the root function 3 { and the polynomial function 1 + {3 are continuous on R. By part 4 of Theorem 4, the
product J({) = 3 { 1 + {3 is continuous on its domain, R.
21. By Theorem 7, the exponential function h5w and the trigonometric function cos 2w are continuous on (> ).
By part 4 of Theorem 4, O(w) = h5w cos 2w is continuous on (> ).
22. By Theorem 7, the trigonometric function sin { and the polynomial function { + 1 are continuous on R.
By part 5 of Theorem 4, k({) =
sin {
is continuous on its domain, {{ | { 6= 1}.
{+1
23. By Theorem 5, the polynomial w4 1 is continuous on (> ). By Theorem 7, ln { is continuous on its domain, (0> ).
By Theorem 9, ln w4 1 is continuous on its domain, which is
w | w4 1 A 0 = w | w4 A 1 = {w | |w| A 1} = (> 1)
(1> )
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VE
24. The sine and cosine functions are continuous everywhere by Theorem 7, so I ({) = sin(cos(sin {)), which is the composite
of sine, cosine, and (once again) sine, is continuous everywhere by Theorem 9.
25. The function | =
1
is discontinuous at { = 0 because the
1 + h1@{
left- and right-hand limits at { = 0 are different.
26. The function | = tan2 { is discontinuous at { = 2 + n, where n is
any integer. The function | = ln tan2 { is also discontinuous
where tan2 { is 0, that is, at { = n. So | = ln tan2 { is
discontinuous at { = 2 q, q any integer.
27. Because we are dealing with root functions, 5 +
{ is continuous on [0> ), { + 5 is continuous on [5> ), so the
5+ {
is continuous on [0> ). Since i is continuous at { = 4, lim i ({) = i (4) = 73 .
quotient i ({) =
{4
5+{
28. Because { is continuous on R, sin { is continuous on R, and { + sin { is continuous on R, the composite function
i ({) = sin({ + sin {) is continuous on R, so lim i ({) = i () = sin( + sin ) = sin = 0.
{
2
29. Because {2 { is continuous on R, the composite function i ({) = h{ { is continuous on R, so
lim i ({) = i (1) = h1 1 = h0 = 1.
{1
30. {3 3{ + 1 = 0 for three values of {, but 2 is not one of them. Thus, i ({) = ({3 3{ + 1)3 is continuous at { = 2 and
1
lim i ({) = i (2) = (8 6 + 1)3 = 33 = 27
.
{2
+ 2
{
31. i ({) =
{
if { ? 1
if { 1
By Theorem 5, since i ({) equals the polynomial {2 on (> 1), i is continuous on (> 1). By Theorem 7, since i ({)
equals the root function
lim i ({) = lim
{1+
{1+
{ on (1> )> i is continuous on (1> ). At { = 1, lim i({) = lim {2 = 1 and
{1
{1
{ = 1. Thus, lim i({) exists and equals 1. Also, i (1) = 1 = 1. Thus, i is continuous at { = 1.
{1
We conclude that i is continuous on (> ).
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SECT
+
32. i ({) =
sin {
if { ? @4
cos {
if { @4
ยค
109
By Theorem 7, the trigonometric functions are continuous. Since i({) = sin { on (> @4) and i ({) = cos { on
(@4> ), i is continuous on (> @4) (@4> )=
function is continuous at @4= Similarly,
@4. Thus,
lim
{(@4)+
lim
{(@4)
i ({) =
i ({) =
lim
{(@4)+
lim
{(@4)
sin { = sin 4 = 1@ 2 since the sine
cos { = 1@ 2 by continuity of the cosine function at
i ({) exists and equals 1@ 2, which agrees with the value i (@4). Therefore, i is continuous at @4, so
lim
{(@4)
i is continuous on (> ).
;
{+2
A
A
?
33. i ({) = h{
A
A
=
2{
if { ? 0
if 0 { 1
if { A 1
i 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 i ({) = lim ({ + 2) = 2 and lim i ({) = lim h{ = 1, so i is discontinuous at 0. Since i (0) = 1, i is
{0
{0
{0+
{0+
continuous from the right at 0. Also lim i ({) = lim h{ = h and lim i ({) = lim (2 {) = 1, so i is discontinuous
{1
{1
{1+
{1+
at 1. Since i (1) = h, i is continuous from the left at 1.
34. By Theorem 5, each piece of I is continuous on its domain. We need to check for continuity at u = U.
lim I (u) = lim
uU
uU
JP
JP
JP
JP
JPu
JP
=
and lim I (u) = lim
=
, so lim I (u) =
. Since I (U) =
,
3
2
2
2
2
+
+
uU
U
U
u
U
U
U2
uU
uU
I is continuous at U. Therefore, I is a continuous function of u.
+
35. i ({) =
f{2 + 2{
3
{ f{
if { ? 2
if { 2
i is continuous on (> 2) and (2> ). Now lim i ({) = lim
{2
lim i ({) = lim
{2+
{2+
{2
2
f{ + 2{ = 4f + 4 and
3
{ f{ = 8 2f. So i is continuous 4f + 4 = 8 2f 6f = 4 f = 23 . Thus, for i
to be continuous on (> ), f = 23 .
36. i ({) =
; 2
{ 4
A
A
A {2
?
2
d{ e{ + 3
A
A
A
=
2{ d + e
At { = 2:
if { ? 2
if 2 { ? 3
if { 3
{2 4
({ + 2)({ 2)
= lim
= lim ({ + 2) = 2 + 2 = 4
{2
{2 { 2
{2
{2
lim i ({) = lim
{2
lim i ({) = lim (d{2 e{ + 3) = 4d 2e + 3
{2+
{2+
( )
We must have 4d e + 3 = 4,, or 4d 2e = 1 (1).
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At { = 3:
LIMITS AND
ND DERIVATIVES
VE
lim i ({) = lim (d{2 e{ + 3) = 9d 3e + 3
{3
{3
lim i ({) = lim (2{ d + e) = 6 d + e
{3+
{3+
We must have 9d 3e + 3 = 6 d + e, or 10d 4e = 3 (2).
Now solve the system of equations by adding 2 times equation (1) to equation (2).
8d + 4e = 2
10d 4e =
2d
=
3
1
So d = 12 . Substituting 12 for d in (1) gives us 2e = 1, so e = 12 as well. Thus, for i to be continuous on (> ),
d = e = 12 .
37. (a) i ({) =
{4 1
({2 + 1)({2 1)
({2 + 1)({ + 1)({ 1)
=
=
= ({2 + 1)({ + 1) [or {3 + {2 + { + 1]
{1
{1
{1
for { 6= 1. The discontinuity is removable and j({) = {3 + {2 + { + 1 agrees with i for { 6= 1 and is continuous on R.
(b) i ({) =
{({2 { 2)
{({ 2)({ + 1)
{3 {2 2{
=
=
= {({ + 1) [or {2 + {] for { 6= 2. The discontinuity
{2
{2
{2
is removable and j({) = {2 + { agrees with i for { 6= 2 and is continuous on R.
(c) lim i ({) = lim [[sin {]] = lim 0 = 0 and lim i ({) = lim [[sin {]] = lim (1) = 1, so lim i ({) does not
{
{
{
{ +
{ +
{ +
{
exist. The discontinuity at { = is a jump discontinuity.
38.
i does not satisfy the conclusion of the
i does satisfy the conclusion of the
Intermediate Value Theorem.
Intermediate Value Theorem.
39. i ({) = {2 + 10 sin { is continuous on the interval [31> 32], i (31)
957, and i(32)
1030. Since 957 ? 1000 ? 1030,
there is a number c in (31> 32) such that i (f) = 1000 by the Intermediate Value Theorem. Note: There is also a number c in
(32> 31) such that i (f) = 1000=
40. Suppose that i(3) ? 6. By the Intermediate Value Theorem applied to the continuous function i on the closed interval [2> 3],
the fact that i (2) = 8 A 6 and i(3) ? 6 implies that there is a number f in (2> 3) such that i (f) = 6. This contradicts the fact
that the only solutions of the equation i ({) = 6 are { = 1 and { = 4. Hence, our supposition that i (3) ? 6 was incorrect. It
follows that i(3) 6. But i (3) 6= 6 because the only solutions of i ({) = 6 are { = 1 and { = 4. Therefore, i (3) A 6.
41. i ({) = {4 + { 3 is continuous on the interval [1> 2]> i (1) = 1, and i(2) = 15. Since 1 ? 0 ? 15, there is a number f
in (1> 2) such that i (f) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation {4 + { 3 = 0 in the
interval (1> 2)=
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111
3
{ + { 1 is continuous on the interval [0> 1]> i (0) = 1, and i (1) = 1. Since 1 ? 0 ? 1, there is a number f in
3
(0> 1) such that i(f) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation { + { 1 = 0, or
3
{ = 1 {, in the interval (0> 1)=
42. i ({) =
43. The equation h{ = 3 2{ is equivalent to the equation h{ + 2{ 3 = 0. i ({) = h{ + 2{ 3 is continuous on the interval
[0> 1], i (0) = 2, and i (1) = h 1
1=72. Since 2 ? 0 ? h 1, there is a number f in (0> 1) such that i (f) = 0 by the
Intermediate Value Theorem. Thus, there is a root of the equation h{ + 2{ 3 = 0, or h{ = 3 2{, in the interval (0> 1).
44. The equation sin { = {2 { is equivalent to the equation sin { {2 + { = 0. i ({) = sin { {2 + { is continuous on the
interval [1> 2]> i (1) = sin 1
0=84, and i (2) = sin 2 2
1=09. Since sin 1 A 0 A sin 2 2, there is a number f in
(1> 2) such that i (f) = 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).
45. (a) i ({) = cos { {3 is continuous on the interval [0> 1], i (0) = 1 A 0, and i (1) = cos 1 1
0=46 ? 0. Since
1 A 0 A 0=46, there is a number f in (0> 1) such that i (f) = 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).
(b) i (0=86)
0=016 A 0 and i (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).
46. (a) i ({) = ln { 3 + 2{ is continuous on the interval [1> 2], i (1) = 1 ? 0, and i (2) = ln 2 + 1
1=7 A 0. Since
1 ? 0 ? 1=7, there is a number f in (1> 2) such that i (f) = 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) i (1=34)
0=03 ? 0 and i (1=35)
0=0001 A 0, so there is a root between 1=34 and 1=35> that is, in the
interval (1=34> 1=35).
47. (a) Let i ({) = 100h{@100 0=01{2 = Then i (0) = 100 A 0 and
i (100) = 100h1 100
63=2 ? 0. So by the Intermediate
Value Theorem, there is a number f in (0> 100) such that i (f) = 0.
This implies that 100hf@100 = 0=01f2 .
(b) Using the intersect feature of the graphing device, we nd that the
root of the equation is { = 70=347, correct to three decimal places.
{5
1
. Then i(5) = 18 ? 0 and i (6) = 89 A 0, and i is continuous on [5> ). So by the
{+3
1
= f 5.
Intermediate Value Theorem, there is a number f in (5> 6) such that i (f) = 0. This implies that
f+3
48. (a) Let i ({) =
(b) Using the intersect feature of the graphing device, we nd
that the root of the equation is { = 5=016, correct to three
decimal places.
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VE
49. lim sin(d + k) = lim (sin d cos k + cos d sin k) = lim (sin d cos k) + lim (cos d sin k)
k0
k0
=
k0
k0
lim sin d lim cos k + lim cos d lim sin k = (sin d)(1) + (cos d)(0) = sin d
k0
k0
k0
k0
50. As in the previous exercise, we must show that lim cos(d + k) = cos d to prove that the cosine function is continuous.
k0
lim cos(d + k) = lim (cos d cos k sin d sin k) = lim (cos d cos k) lim (sin d sin k)
k0
k0
=
k0
k0
lim cos d lim cos k lim sin d lim sin k = (cos d)(1) (sin d)(0) = cos d
k0
k0
k0
51. If there is such a number, it satises the equation {3 + 1 = {
k0
{3 { + 1 = 0. Let the left-hand side of this equation be
called i ({). Now i (2) = 5 ? 0, and i (1) = 1 A 0. Note also that i ({) is a polynomial, and thus continuous. So by the
Intermediate Value Theorem, there is a number f between 2 and 1 such that i(f) = 0, so that f = f3 + 1.
52.
d
{3 + 2{2 1
+
e
{3 + { 2
= 0 d({3 + { 2) + e({3 + 2{2 1) = 0. Let s({) denote the left side of the last
equation. Since s is continuous on [1> 1], s(1) = 4d ? 0, and s(1) = 2e A 0, there exists a f in (1> 1) such that
s(f) = 0 by the Intermediate Value Theorem. Note that the only root of either denominator that is in (1> 1) is
(1 + 5 )@2 = u, but s(u) = (3 5 9)d@2 6= 0. Thus, f is not a root of either denominator, so s(f) = 0
{ = f is a root of the given equation.
53. i ({) = {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 i(0). Thus, i is
{0
{0
{0
continuous at 0 and, hence, on (> ).
54. (a) lim I ({) = 0 and lim I ({) = 0, so lim I ({) = 0, which is I (0), and hence I is continuous at { = d if d = 0. For
{0+
{0
{0
d A 0, lim I ({) = lim { = d = I (d). For d ? 0, lim I ({) = lim ({) = d = I (d). Thus, I is continuous at
{d
{d
{d
{ = d; that is, continuous everywhere.
(b) Assume that i is continuous on the interval L. Then for d
{d
L, lim |i({)| = lim i ({) = |i (d)| by Theorem 8. (If d is
{d
{d
an endpoint of L, use the appropriate one-sided limit.) So |i | is continuous on L.
+
1 if { 0
(c) No, the converse is false. For example, the function i({) =
is not continuous at { = 0, but |i({)| = 1 is
1 if { ? 0
continuous on R.
55. Dene x(w) to be the monkโs distance from the monastery, as a function of time, on the rst day, and dene g(w) to be his
distance from the monastery, as a function of time, on the second day. Let G be the distance from the monastery to the top of
the mountain. From the given information we know that x(0) = 0, x(12) = G, g(0) = G and g(12) = 0. Now consider the
function x g, which is clearly continuous. We calculate that (x g)(0) = G and (x g)(12) = G. So by the
Intermediate Value Theorem, there must be some time w0 between 0 and 12 such that (x g)(w0 ) = 0 x(w0 ) = g(w0 ).
So at time w0 after 7:00 AM, the monk will be at the same place on both days.
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LIMIT INVOLVING INFINITY
ยค
113
2.5 Limits Involving Infinity
1. (a) As { approaches 2 (from the right or the left), the values of i ({) become large.
(b) As { approaches 1 from the right, the values of i ({) become large negative.
(c) As { becomes large, the values of i ({) approach 5.
(d) As { becomes large negative, the values of i ({) 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 innite
meet but not cross it.
number of times.
(b) The graph of a function can have 0, 1, or 2 horizontal asymptotes. Representative examples are shown.
No horizontal asymptote
3. (a) lim i ({) =
{2
(e) lim i ({) = 2
{
4. (a) lim j({) = 2
{
(d) lim j({) =
{0
5. lim i ({) = ,
{0
lim i({) = 5,
{
lim i ({) = 5
{
(b)
One horizontal asymptote
lim i({) =
(c)
{1
Two horizontal asymptotes
lim i ({) =
(d) lim i ({) = 1
{
{1+
(f ) Vertical: { = 1, { = 2; Horizontal: | = 1, | = 2
(b) lim j({) = 2
(c) lim j({) =
{
(e)
{3
lim j({) =
(f ) Vertical: { = 2, { = 0, { = 3; Horizontal: | = 2, | = 2
{2+
6. lim i ({) = ,
{2
lim i ({) = ,
{2+
lim i ({) = ,
{2
lim i ({) = 0,
{
lim i({) = 0,
{
i (0) = 0
7. lim i ({) = ,
{2
lim i ({) = 0,
{
lim i ({) = ,
{
lim i ({) = ,
{0+
lim i ({) =
{0
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CHAPTER 2 LIMITS AND
D DERIVATIVES
VE
8. . lim i ({) = ,
9. i (0) = 3,
{3+
lim i ({) = ,
lim i ({) = 4,
10. lim i ({) = ,
{3
{0
i (0) = 0, i is even
lim i ({) = 2,
{3
{0+
lim i ({) = > i is odd
{
lim i({) = ,
{
lim i ({) = ,
{4+
lim i ({) = 2,
{
lim i({) = ,
{4
lim i({) = 3
{
11. If i ({) = {2@2{ , then a calculator gives i(0) = 0, i (1) = 0=5, i(2) = 1, i (3) = 1=125, i (4) = 1, i (5) = 0=78125,
i (6) = 0=5625, i (7) = 0=3828125, i (8) = 0=25, i (9) = 0=158203125, i(10) = 0=09765625, i (20)
2=2204 ร 1012 , i (100)
It appears that lim {2@2{ = 0.
i (50)
0=00038147,
7=8886 ร 1027 .
{
12. (a) i ({) =
1
.
{3 1
{
i ({)
{
i ({)
0=5
1=14
1=5
0=42
From these calculations, it seems that
0=9
3=69
1=1
3=02
lim i ({) = and lim i ({) = .
0=99
33=7
1=01
33=0
0=999
333=7
1=001
333=0
0=9999
3333=7
1=0001
3333=0
0=99999
33,333=7
1=00001
33,333=3
{1
{1+
(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, i ({),
will be a negative number with large absolute value. So lim i({) = .
{1
If { is slightly larger than 1, then {3 1 will be a small positive number, and its reciprocal, i ({), will be a large positive
number. So lim i ({) = .
{1+
(c) It appears from the graph of i that
lim i ({) = and lim i ({) = .
{1
{1+
INSTRUCTOR USE ONLY
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SECTION 22.55 LIMITS
LIMIT INVOLVING INFINITY
13. Vertical: {
1=62, {
ยค
115
0=62, { = 1;
Horizontal: | = 1
{
14. (a) From a graph of i({) = (1 2@{) in a window of [0> 10,000] by [0> 0=2], we estimate that lim i ({) = 0=14
{
(to two decimal places.)
(b)
15. lim
10,000
100,000
1,000,000
0=135308
0=135333
0=135335
2{
lim
{3
= since the numerator is positive and the denominator approaches 0 through positive values as {
{+2
= since the numerator is negative and the denominator approaches 0 from the negative side as {
{+3
17. Let w = 3@(2 {). As {
18. lim cot { = lim
{
as {
{
i ({)
{1 ({ 1)2
16.
From the table, we estimate that lim i ({) = 0=1353 (to four decimal places.)
{
{
2+ , w
1.
3 .
. So lim h3@(2{) = lim hw = 0 by (7).
{2+
w
cos {
= since the numerator is negative and the denominator approaches 0 through positive values
sin {
.
19. Let w = {2 9. Then as {
3+ , w
0+ , and lim ln({2 9) = lim ln w = by (3).
{3+
w0+
{({ 2)
{2 2{
{
= lim
= lim
= since the numerator is positive and the denominator
{2 {2 4{ + 4
{2 ({ 2)2
{2 { 2
20. lim
approaches 0 through negative values as {
21.
lim { csc { = lim
{2
values as {
{
{2 sin {
2 .
= since the numerator is positive and the denominator approaches 0 through negative
2 .
1
lim 3 + 5 lim
3{ + 5
(3{ + 5)@{
3 + 5@{
3 + 5(0)
{
{ {
=
= lim
= lim
=
=3
22. lim
1
{ { 4
{ ({ 4)@{
{ 1 4@{
1 4(0)
lim 1 4 lim
{
{ {
23. Divide both the numerator and denominator by {3 (the highest power of { that occurs in the denominator).
5
{3 + 5{
5
lim 1 + 2
1+ 2
{
{
{3 + 5{
{3
{
=
lim
=
lim
=
lim
1
4
{ 2{3 {2 + 4
{ 2{3 {2 + 4
{
4
1
2 + 3
lim 2 + 3
{ {
{
{
{
{3
1
lim 1 + 5 lim 2
1 + 5(0)
1
{
{ {
=
=
=
1
1
2 0 + 4(0)
2
+ 4 lim 3
lim 2 lim
{
{
{
{
{
{
{
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ยค
CHAPTER 2 LIMITS AND
D DERIVATIVES
VE
2
w + 2 @w3
w2 + 2
1@w + 2@w3
0+0
=
lim
=0
24. lim 3
=
lim
=
w w + w2 1
w (w3 + w2 1) @w3
w 1 + 1@w 1@w3
1+00
25. First, multiply the factors in the denominator. Then divide both the numerator and denominator by x4 .
5
4×4 + 5
4+ 4
4
4×4 + 5
4×4 + 5
x
x
lim
= lim
= lim
= lim
5
2
x (x2 2)(2×2 1)
x 2×4 5×2 + 2
x 2×4 5×2 + 2
x
2 2 + 4
4
x
x
x
5
1
lim 4 + 4
lim 4 + 5 lim 4
x
x
4
4 + 5(0)
x
x x
=
= =2
=
=
1
1
2 5(0) + 2(0)
2
2
5
lim 2 5 lim 2 + 2 lim 4
lim 2 2 + 4
x
x x
x x
x
x
x
{+2
({ + 2) @{
1 + 2@{
1
1+0
= lim s
=
=
= lim
{
3
9+0
9{2 + 1 { 9{2 + 1@ {2
9 + 1@{2
26. lim
{
2
9{2 + { 3{
9{2 + { + 3{
9{2 + { (3{)2
= lim
{
{
9{2 + { + 3{
9{2 + { + 3{
2
9{ + { 9{2
{
1@{
= lim
ยท
= lim
2
2
{
{
9{ + { + 3{
9{ + { + 3{ 1@{
1
{@{
1
1
1
= lim s
=
= lim s
=
=
2
2
2
{
{
3+3
6
9+3
9{ @{ + {@{ + 3{@{
9 + 1@{ + 3
27. lim
9{2 + { 3{ = lim
{
{2 + d{ {2 + e{
{2 + d{ + {2 + e{
{
{2 + d{ + {2 + e{
28. lim
{2 + d{ {2 + e{ = lim
{
({2 + d{) ({2 + e{)
[(d e){]@{
= lim
= lim
2
2
2
{
{
{ + d{ + { + e{
{ + d{ + {2 + e{ @ {2
de
de
de
s
=
= lim s
=
{
2
1
+
0
+
1
+
0
1 + d@{ + 1 + e@{
29. Let w = {2 . As {
30. For { A 0,
, w
2
. So lim h{ = lim hw = 0 by (7).
{
{2 + 1 A {2 = {. So as {
w
, we have
{2 + 1
, that is, lim
{
{2 + 1 = .
31. lim cos { does not exist because as { increases cos { does not approach any one value, but oscillates between 1 and 1.
{
32. Since 0 sin2 { 1, we have 0
sin2 {
1
1
2 . Now lim 0 = 0 and lim 2 = 0, so by the Squeeze Theorem,
{
{ {
{2
{
sin2 {
= 0.
{
{2
lim
33. Since 1 cos { 1 and h2{ A 0, we have h2{ h2{ cos { h2{ . We know that lim (h2{ ) = 0 and
lim h2{ = 0, so by the Squeeze Theorem, lim (h2{ cos {) = 0.
{
{
{
h3{ h3{
1 h6{
10
=1
=
lim
=
{ h3{ + h3{
{ 1 + h6{
1+0
34. Divide numerator and denominator by h3{ : lim
35.
lim ({4 + {5 ) = lim {5 ( {1 + 1) [factor out the largest power of {] = because {5
{
as {
Or:
{
.
lim {4 + {5 = lim {4 (1 + {) = .
{
{
and 1@{ + 1
1
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SECTION 22.5 LIMITS
LIMIT INVOLVING INFINITY
36. If we let w = tan {, then as {
(@2)+ , w
. Thus,
{ + {3 + {5
({ + {3 + {5 )@{4
= lim
2
4
{ 1 { + {
{ (1 {2 + {4 )@{4
lim
{(@2)+
ยค
htan { = lim hw = 0.
w
[divide by the highest power of { in the denominator]
37. lim
1@{3 + 1@{ + {
=
{ 1@{4 1@{2 + 1
= lim
because (1@{3 + 1@{ + {)
and (1@{4 1@{2 + 1)
1 as {
.
38. (a)
From the graph, it appears at rst 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 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
lim
{ 3{ 5
2{2 + 1
0=5
and
lim
0=5
{ 3{ 5
2{2 + 1
0=47.
(b) i (1000) 0=4722 and i(10,000) 0=4715, so we estimate that lim
{ 3{ 5
2{2 + 1
i (1000) 0=4706 and i(10,000) 0=4713, so we estimate that lim
0=47.
{ 3{ 5
s
2 + 1@{2
2{2 + 1
2
2
= lim
[since { = { for { A 0] =
0=471404.
(c) lim
{ 3{ 5
{
3 5@{
3
For { ? 0, we have {2 = |{| = {, so when we divide the numerator by {, with { ? 0, we
s
1 2
1 2
2{ + 1 =
2{ + 1 = 2 + 1@{2 . Therefore,
{
{2
s
2 + 1@{2
2{2 + 1
2
= lim
=
0=471404.
lim
{ 3{ 5
{
3 5@{
3
get
1
1
2{2 + { 1
1
1
lim 2 + 2
2+ 2
{
{
2{2 + { 1
{2
{
{ = {
=
lim
39. lim
=
lim
1
2
{ {2 + { 2
{ {2 + { 2
{
2
1
1+ 2
lim
1
+
{
{
{
{
{2
{2
lim 2 + lim
=
| = i ({) =
{
1
{ {
lim
1
{ {2
1
1
lim 1 + lim
2 lim 2
{
{ {
{ {
=
2+00
= 2, so | = 2 is a horizontal asymptote.
1 + 0 2(0)
2{2 + { 1
(2{ 1)({ + 1)
=
, so lim i({) = ,
{2 + { 2
({ + 2)({ 1)
{2
lim i ({) = , lim i ({) = , and lim i ({) = . Thus, { = 2
{2+
{1
{1+
and { = 1 are vertical asymptotes. The graph conrms our work.
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CHAPTER 2 LIMITS AND
D DERIVATIVES
VE
1
{2 + 1
1
lim 1 + 2
1+ 2
{
{
{2 + 1
{2
{
=
lim
40. lim
=
lim
=
3
2
{ 2{2 3{ 2
{ 2{2 3{ 2
{
2
3
2 2
lim
2
{
{
{
{ {2
{2
1
lim 1 + lim
{
=
{ {2
3
2
lim 2 lim
lim 2
{
{ {
{ {
=
1
1+0
= , so | = 12 is a horizontal asymptote.
200
2
| = i ({) =
{2 + 1
{2 + 1
=
, so
lim
i ({) =
2{2 3{ 2
(2{ + 1)({ 2)
{(1@2)
because as {
(1@2) the numerator is positive while the denominator
approaches 0 through positive values. Similarly,
lim
{(1@2)+
i({) = ,
lim i ({) = , and lim i({) = . Thus, { = 12 and { = 2 are vertical
{2
{2+
asymptotes. The graph conrms our work.
41. | = i ({) =
{3 {
{2 6{ + 5
=
{({2 1)
{({ + 1)({ 1)
{({ + 1)
=
=
= j({) for { 6= 1.
({ 1)({ 5)
({ 1)({ 5)
{5
The graph of j is the same as the graph of i with the exception of a hole in the
graph of i at { = 1. By long division, j({) =
As {
ยฑ, j({)
{2 + {
30
={+6+
.
{5
{5
ยฑ, so there is no horizontal asymptote. The denominator
of j is zero when { = 5. lim j({) = and lim j({) = , so { = 5 is a
{5
{5+
vertical asymptote. The graph conrms our work.
42. lim
2h{
{ h{ 5
= lim
2h{
2h{
{ h{ 5
ยท
1@h{
2
2
= lim
=
= 2, so | = 2 is a horizontal asymptote.
{ 1 (5@h{ )
1@h{
10
2(0)
= 0, so | = 0 is a horizontal asymptote. The denominator is zero (and the numerator isnโt)
05
when h{ 5 = 0 h{ = 5 { = ln 5.
lim
{ h{ 5
=
2h{
= since the numerator approaches 10 and the denominator
{(ln 5)+ h{ 5
lim
approaches 0 through positive values as {
lim
2h{
{(ln 5) h{ 5
(ln 5)+ . Similarly,
= . Thus, { = ln 5 is a vertical asymptote. The graph
conrms our work.
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SECTION 22.5 LIMITS
LIMIT INVOLVING INFINITY
(b)
43. (a)
From the graph of i ({) =
{2 + { + 1 + {, we
{
i({)
10,000
0=4999625
100,000
0=4999962
1,000,000
0=4999996
ยค
From the table, we estimate the limit to be 0=5.
estimate the value of lim i ({) to be 0=5.
{
2
2
{ + { + 1 {2
{ +{+1{
2
2
(c) lim
{ + { + 1 + { = lim
{ +{+1+{
= lim
{
{
{
{2 + { + 1 {
{2 + { + 1 {
({ + 1)(1@{)
1 + (1@{)
s
= lim
= lim
2
{
{
{ + { + 1 { (1@{)
1 + (1@{) + (1@{2 ) 1
1+0
1
=
=
2
1+0+01
Note that for { ? 0, we have
{2 = |{| = {, so when we divide the radical by {, with { ? 0, we get
s
1 2
1 2
{ + { + 1 =
{ + { + 1 = 1 + (1@{) + (1@{2 ).
{
{2
(b)
44. (a)
{
i({)
10,000
1=44339
100,000
1=44338
1,000,000
1=44338
From the table, we estimate (to four decimal places)
From the graph of
i ({) = 3{2 + 8{ + 6 3{2 + 3{ + 1, we
the limit to be 1=4434.
estimate (to one decimal place) the value of lim i ({)
{
to be 1=4.
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
2
2
2
{
{
3{ + 8{ + 6 + 3{ + 3{ + 1
3{ + 8{ + 6 + 3{2 + 3{ + 1 (1@{)
5 + 5@{
5
5
5 3
s
= =
1=443376
= lim s
=
{
6
3+ 3
2 3
3 + 8@{ + 6@{2 + 3 + 3@{ + 1@{2
(c) lim i ({) = lim
{
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CHAPTER 2 LIMITS AND
ND DERIVATIVES
VE
45. From the graph, it appears | = 1 is a horizontal asymptote.
3{3 + 500{2
3{ + 500{
3 + (500@{)
{3
= lim 3
lim
= lim
{ {3 + 500{2 + 100{ + 2000
{ { + 500{2 + 100{ + 2000
{ 1 + (500@{) + (100@{2 ) + (2000@{3 )
{3
3+0
= 3, so | = 3 is a horizontal asymptote.
=
1+0+0+0
3
2
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.
46. (a)
No, because the calculator-produced graph of i ({) = h{ + ln |{ 4| looks like an exponential function, but the graph of i
has an innite discontinuity at { = 4. A second graph, obtained by increasing the numpoints option in Maple, begins to
reveal the discontinuity at { = 4.
(b) There isnโt a single graph that shows all the features of i . Several graphs are needed since i looks like ln |{ 4| for large
negative values of { and like h{ for { A 5, but yet has the innite discontiuity at { = 4.
A hand-drawn graph, though distorted, might be better at revealing the main
features of this function.
47. Letโs look for a rational function.
(1)
lim i({) = 0 degree of numerator ? degree of denominator
{ยฑ
(2) lim i ({) = 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 i ({) = and lim i({) = vertical asymptote at { = 3; there is a factor of ({ 3) in the
INSTRUCTOR USE ONLY
{3
{
3
3
{3
{
3
3+
denominator.
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SECTION 22.55 LIMITS
LIMIT INVOLVING INFINITY
ยค
121
(4) i (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
i ({) =
2{
as one possibility.
{2 ({ 3)
48. 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 coefcients must be 1. One possibility is i ({) =
{2
.
({ 1)({ 3)
49. (a) We must rst nd the function i . Since i 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, i now looks like this: i ({) =
d({ 1)({ + 1)
d({ 1)
d(1 1)
2
2
= lim
=
= d, so d = 2, and
{1 { 4
({ 4)({ + 1)
(1 4)
5
5
be determined. Then lim i ({) = lim
{1
d = 5. Thus i ({) =
i (0) =
d({ 1)({ + 1)
, where d is still to
({ 4)({ + 1)
{1
5({ 1)({ + 1)
is a ratio of quadratic functions satisfying all the given conditions and
({ 4)({ + 1)
5(1)(1)
5
= .
(4)(1)
4
{2 1
({2 @{2 ) (1@{2 )
10
=
5
lim
=5
= 5(1) = 5
{ {2 3{ 4
{ ({2 @{2 ) (3{@{2 ) (4@{2 )
100
(b) lim i ({) = 5 lim
{
50. (a) In both viewing rectangles,
lim S ({) = lim T({) = and
{
{
lim S ({) = lim T({) = .
{
{
In the larger viewing rectangle, S and T
become less distinguishable.
2 1
5 1
S ({)
3{5 5{3 + 2{
(b) lim
= lim 1 ยท 2 + ยท 4 = 1 53 (0) + 23 (0) = 1
= lim
{ T({)
{
{
3{5
3 {
3 {
S and T have the same end behavior.
51. (a) Divide the numerator and the denominator by the highest power of { in T({).
(a) If deg S ? deg T, then the numerator
(b) If deg S A deg T, then the numerator
0 but the denominator doesnโt. So lim [S ({)@T({)] = 0.
{
ยฑ but the denominator doesnโt, so lim [S ({)@T({)] = ยฑ
{
(depending on the ratio of the leading coefcients of S and T).
52.
(i) q = 0
(ii) q A 0 (q odd)
(iii) q A 0 (q even)
(iv) q ? 0 (q odd)
(v) q ? 0 (q even)
From these sketches we see that
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NOT FOR SALE
ยค
CHAPTER 2 LIMITS AND
ND DERIVATIVES
VE
;
1 if q = 0
A
?
q
(a) lim { = 0 if q A 0
A
{0+
=
if q ? 0
(b) lim {q =
;
A
A
A
A
?
1 if q = 0
0 if q A 0
A
if q ? 0, q odd
A
A
A
=
if q ? 0, q even
;
1 if q = 0
A
A
A
A
? if q A 0, q odd
(d) lim {q =
{
A
A
A if q A 0, q even
A
=
0 if q ? 0
{0
;
1 if q = 0
A
?
(c) lim {q = if q A 0
{
A
=
0 if q ? 0
5 {
5
5
1@ {
=
ยท = lim s
= 5 and
{
{ 1 1@ {
10
1 (1@{)
53. lim
{
10h{ 21 1@h{
10 (21@h{ )
10 0
10h{ 21
5 {
=
=
5.
Since
,
ยท
=
lim
?
i
({)
?
{
{
2h{
1@h{
2
2
2h{
{1
lim
we have lim i({) = 5 by the Squeeze Theorem.
{
p0
. As y
1 y2@f2
54. lim p = lim s
yf
yf
f ,
s
1 y 2@f2
0+ , and p
.
55. (a) After w minutes, 25w liters of brine with 30 g of salt per liter has been pumped into the tank, so it contains
(5000 + 25w) liters of water and 25w ยท 30 = 750w grams of salt. Therefore, the salt concentration at time w will be
750w
30w g
F(w) =
=
.
5000 + 25w
200 + w L
(b) lim F(w) = lim
w
30w
w 200 + w
30w@w
= lim
w 200@w + w@w
30
= 30. So the salt concentration approaches that of the brine
0+1
=
being pumped into the tank.
56. (a) lim y(w) = lim y 1 hjw@y
= y (1 0) = y
w
w
(b) We graph y(w) = 1 h9=8w and y(w) = 0=99y , or in this case,
y(w) = 0=99. Using an intersect feature or zooming in on the point of
intersection, we nd that w
0=47 s.
57. (a) If w = {@10, then { = 10w and as {
, w
(b) | = h{@10 and | = 0=1 intersect at {1
23=03.
. Thus, lim h{@10 = lim hw = 0 by Equation 7.
{
w
If { A {1 , then h{@10 ? 0=1.
(c) h{@10 ? 0=1 {@10 ? ln 0=1
1
= 10 ln 101 = 10 ln 10
{ A 10 ln 10
23=03
4 5@{
4
4{2 5{
= =2
= lim
{ 2{2 + 1
{ 2 + 1@{2
2
58. (a) lim i ({) = lim
{
(b) i ({) = 1=9 {
i ({) = 1=99 {
25=3744, so i ({) A 1=9 when { A Q = 25=4.
250=3974, so i ({) A 1=99 when { A Q = 250=4.
INSTRUCTOR USE
SE
S
E ON
ONLY
ON
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SECTION
EC
22.66 DERIVATIVES AND
A RATES OF CHANGE
ยค
123
2.6 Derivatives and Rates of Change
1. (a) This is just the slope of the line through two points: pS T =
i ({) i (3)
|
=
.
{
{3
(b) This is the limit of the slope of the secant line S T as T approaches S : p = lim
{3
i({) i (3)
.
{3
2. The curve looks more like a line as the viewing rectangle gets smaller.
3. (a) (i) Using Denition 1 with i ({) = 4{ {2 and S (1> 3),
i ({) i (d)
(4{ {2 ) 3
({2 4{ + 3)
({ 1)({ 3)
= lim
= lim
= lim
{d
{1
{1
{1
{d
{1
{1
{1
p = lim
= lim (3 {) = 3 1 = 2
{1
(ii) Using Equation 2 with i ({) = 4{ {2 and S (1> 3),
4(1 + k) (1 + k)2 3
i (d + k) i(d)
i (1 + k) i (1)
= lim
= lim
p = lim
k0
k0
k0
k
k
k
4 + 4k 1 2k k2 3
k2 + 2k
k(k + 2)
= lim
= lim
= lim (k + 2) = 2
k0
k0
k0
k0
k
k
k
= lim
(b) An equation of the tangent line is | i(d) = i 0 (d)({ d) | i (1) = i 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 Denition 1 with i ({) = { {3 and S (1> 0),
i ({) 0
{ {3
{(1 {2 )
{(1 + {)(1 {)
= lim
= lim
= lim
{1 { 1
{1 { 1
{1
{1
{1
{1
p = lim
= lim [{(1 + {)] = 1(2) = 2
{1
(ii) Using Equation 2 with i ({) = { {3 and S (1> 0),
(1 + k) (1 + k)3 0
i (d + k) i (d)
i(1 + k) i(1)
p = lim
= lim
= lim
k0
k0
k0
k
k
k
1 + k (1 + 3k + 3k2 + k3 )
k(k2 3k 2)
k3 3k2 2k
= lim
= lim
k0
k0
k0
k
k
k
= lim
= lim (k2 3k 2) = 2
k0
INSTRUCTOR USE ONLY
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CHAPTER 2
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LIMITS AND
ND DERIVATIVES
VE
(b) An equation of the tangent line is | i (d) = i 0 (d)({ d) | i (1) = i 0 (1)({ 1) | 0 = 2({ 1),
or | = 2{ + 2.
The graph of | = 2{ + 2 is tangent to the graph of | = { {3 at the
(c)
point (1> 0). Now zoom in toward the point (1> 0) until the cubic and the
tangent line are indistinguishable.
5. Using (1) with i ({) = 4{ 3{2 and S (2> 4) [we could also use (2)],
4{ 3{2 (4)
i ({) i (d)
3{2 + 4{ + 4
p = lim
= lim
= lim
{d
{2
{2
{d
{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 i ({) = {3 3{ + 1 and S (2> 3),
i (d + k) i (d)
i (2 + k) i (2)
(2 + k)3 3(2 + k) + 1 3
= lim
= lim
k0
k0
k0
k
k
k
p = lim
8 + 12k + 6k2 + k3 6 3k 2
9k + 6k2 + k3
k(9 + 6k + k2 )
= lim
= lim
k0
k0
k0
k
k
k
= lim
= lim (9 + 6k + k2 ) = 9
k0
Tangent line: | 3 = 9({ 2) | 3 = 9{ 18 | = 9{ 15
( { 1)( { + 1)
{ 1
{1
1
1
7. Using (1), p = lim
= lim
= lim
= .
= lim
{1
{1 ({ 1)( { + 1)
{1 ({ 1)( { + 1)
{1
{1
2
{+1
Tangent line: | 1 = 12 ({ 1)
8. Using (1) with i ({) =
| = 12 { + 12
2{ + 1
and S (1> 1),
{+2
2{ + 1
2{ + 1 ({ + 2)
1
i ({) i (d)
{1
{+2
= lim { + 2
= lim
= lim
p = lim
{d
{1
{1
{1
{d
{1
{1
({ 1)({ + 2)
= lim
1
{1 { + 2
=
1
1
=
1+2
3
Tangent line: | 1 = 13 ({ 1)
| 1 = 13 { 13
| = 13 { + 23
9. (a) Using (2) with | = i ({) = 3 + 4{2 2{3 ,
i (d + k) i (d)
3 + 4(d + k)2 2(d + k)3 (3 + 4d2 2d3 )
= lim
k0
k0
k
k
p = lim
3 + 4(d2 + 2dk + k2 ) 2(d3 + 3d2 k + 3dk2 + k3 ) 3 4d2 + 2d3
k0
k
= lim
3 + 4d2 + 8dk + 4k2 2d3 6d2 k 6dk2 2k3 3 4d2 + 2d3
k0
k
= lim
8dk + 4k2 6d2 k 6dk2 2k3
k(8d + 4k 6d2 6dk 2k2 )
= lim
k0
k0
k
k
= lim
INSTRUCTOR USE ONLY
= lim (8
(8d
d + 4k
4k 6
6d
d2 6
6dk
dk 2
2k
k2 ) = 8d
8d 6d
6d2
k0
k
0
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SECTION
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22.66 DERIVATIVES AND
A RATES OF CHANGE
(b) At (1> 5): p = 8(1) 6(1)2 = 2, so an equation of the tangent line
ยค
125
(c)
is | 5 = 2({ 1) | = 2{ + 3.
At (2> 3): p = 8(2) 6(2)2 = 8, so an equation of the tangent
line is | 3 = 8({ 2) | = 8{ + 19.
10. (a) Using (1),
1
1
{
d
= lim
p = lim
{d
{d
{d
d {
( d {)( d + {)
d{
d{
= lim
= lim
{d
{d
d{ ({ d) ( d + { ) {d d{ ({ d) ( d + { )
1
1
1
1
= lim
=
= 3@2 or d3@2
{d
2
2d
d{ ( d + { )
d2 (2 d )
(b) At (1> 1): p = 12 , so an equation of the tangent line
(c)
is | 1 = 12 ({ 1) | = 12 { + 32 .
1
At 4> 12 : p = 16
, so an equation of the tangent line
1
1
({ 4) | = 16
{ + 34 .
is | 12 = 16
11. (a) The particle is moving to the right when v is increasing; that is, on the intervals (0> 1) and (4> 6). The particle is moving to
the left when v is decreasing; that is, on the interval (2> 3). The particle is standing still when v 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
30
= 3. On the interval (2> 3), the slope is
10
31
13
= 2. On the interval (4> 6), the slope is
= 1.
32
64
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 nishes the race at a faster velocity.
(b) The distance between the runners is the greatest at the time when the largest vertical line segment ts between the two
graphsโthis 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 equalโthis 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 v(w) = 40w 16w2 .
40w 16w2 16
8 2w2 5w + 2
v(w) v(2)
16w2 + 40w 16
= lim
= lim
= lim
y(2) = lim
w2
w2
w2
w2
w2
w2
w2
w2
= lim
w2
8(w 2)(2w 1)
= 8 lim (2w 1) = 8(3) = 24
w2
w2
INSTRUCTOR USE ONLY
24 ft@s.
f @s.
ft
s.
Thus, the instantaneous velocity when w = 2 is 24
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LIMITS AND
ND DERIVATIVES
VE
14. (a) Let K(w) = 10w 1=86w2 .
10(1 + k) 1=86(1 + k)2 (10 1=86)
K(1 + k) K(1)
= lim
y(1) = lim
k0
k0
k
k
10 + 10k 1=86(1 + 2k + k2 ) 10 + 1=86
k0
k
= lim
10 + 10k 1=86 3=72k 1=86k2 10 + 1=86
k0
k
= lim
6=28k 1=86k2
= lim (6=28 1=86k) = 6=28
k0
k0
k
= lim
The velocity of the rock after one second is 6=28 m@s.
10(d + k) 1=86(d + k)2 (10d 1=86d2 )
K(d + k) K(d)
= lim
(b) y(d) = lim
k0
k0
k
k
10d + 10k 1=86(d2 + 2dk + k2 ) 10d + 1=86d2
k0
k
= lim
10d + 10k 1=86d2 3=72dk 1=86k2 10d + 1=86d2
10k 3=72dk 1=86k2
= lim
k0
k0
k
k
= lim
= lim
k0
k(10 3=72d 1=86k)
= lim (10 3=72d 1=86k) = 10 3=72d
k0
k
The velocity of the rock when w = d is (10 3=72d) m@s=
(c) The rock will hit the surface when K = 0 10w 1=86w2 = 0 w(10 1=86w) = 0 w = 0 or 1=86w = 10.
The rock hits the surface when w = 10@1=86
5=4 s.
10
10
= 10 3=72 1=86
= 10 20 = 10 m@s.
(d) The velocity of the rock when it hits the surface is y 1=86
1
d2 (d + k)2
1
v(d + k) v(d)
(d + k)2
d2
d2 (d + k)2
d2 (d2 + 2dk + k2 )
= lim
= lim
= lim
15. y(d) = lim
k0
k0
k0
k0
k
k
k
kd2 (d + k)2
(2dk + k2 )
k(2d + k)
(2d + k)
2d
2
= lim
= lim 2
= 2 2 = 3 m@s
k0 kd2 (d + k)2
k0 kd2 (d + k)2
k0 d (d + k)2
d ยทd
d
= lim
So y (1) =
2
2
1
2
2
m@s.
= 2 m@s, y(2) = 3 = m@s, and y(3) = 3 =
13
2
4
3
27
16. (a) The average velocity between times w and w + k is
(w + k)2 8(w + k) + 18 w2 8w + 18
v(w + k) v(w)
w2 + 2wk + k2 8w 8k + 18 w2 + 8w 18
=
=
(w + k) w
k
k
=
2wk + k2 8k
= (2w + k 8) m@s.
k
(i) [3> 4]: w = 3, k = 4 3 = 1, so the average
velocity is 2(3) + 1 8 = 1 m@s.
(iii) [4> 5]: w = 4, k = 1, so the average velocity
is 2(4) + 1 8 = 1 m@s.
(ii) [3=5> 4]: w = 3=5, k = 0=5, so the average velocity
is 2(3=5) + 0=5 8 = 0=5 m@s.
(iv) [4> 4=5]: w = 4, k = 0=5, so the average velocity
is 2(4) + 0=5 8 = 0=5 m@s.
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22.66 DERIVATIVES AND
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(b) y(w) = lim
k0
v(w + k) v(w)
= lim (2w + k 8) = 2w 8,
k0
k
ยค
(c)
so y (4) = 0.
17. j 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, j0 (0) ? 0 ? j0 (4) ? j 0 (2) ? j 0 (2).
18. Since j(5) = 3, the point (5> 3) is on the graph of j. Since j0 (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.
19. For the tangent line | = 4{ 5: when { = 2, | = 4(2) 5 = 3 and its slope is 4 (the coefcient of {). At the point of
tangency, these values are shared with the curve | = i ({); that is, i (2) = 3 and i 0 (2) = 4.
20. Since (4> 3) is on | = i ({), i (4) = 3. The slope of the tangent line between (0> 2) and (4> 3) is 14 , so i 0 (4) = 14 .
21. We begin by drawing a curve through the origin with a
slope of 3 to satisfy i (0) = 0 and i 0 (0) = 3. Since
i 0 (1) = 0, we will round off our gure 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.
22. We begin by drawing a curve through the origin with a slope of 1 to satisfy
j(0) = 0 and j 0 (0) = 1. We round off our gure at { = 1 to satisfy j0 (1) = 0,
and then pass through (2> 0) with slope 1 to satisfy j(2) = 0 and j 0 (2) = 1.
We round the gure at { = 3 to satisfy j 0 (3) = 0, and then pass through (4> 0)
with slope 1 to satisfy j(4) = 0 and j 0 (4) = 1= Finally we extend the curve on
both ends to satisfy lim j({) = and lim j({) = .
{
{
23. Using (4) with i ({) = 3{2 {3 and d = 1,
i (1 + k) i (1)
[3(1 + k)2 (1 + k)3 ] 2
= lim
k0
k0
k
k
i 0 (1) = lim
(3 + 6k + 3k2 ) (1 + 3k + 3k2 + k3 ) 2
k(3 k2 )
3k k3
= lim
= lim
k0
k0
k0
k
k
k
= lim
= lim (3 k2 ) = 3 0 = 3
k0
Tangent line: | 2 = 3({ 1) | 2 = 3{ 3 | = 3{ 1
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LIMITS AND
ND DERIVATIVES
VE
24. Using (5) with j({) = {4 2 and d = 1,
j({) j(1)
({4 2) (1)
{4 1
({2 + 1)({2 1)
= lim
= lim
= lim
{1
{1
{1 { 1
{1
{1
{1
{1
j 0 (1) = lim
({2 + 1)({ + 1)({ 1)
= lim [({2 + 1)({ + 1)] = 2(2) = 4
{1
{1
{1
= lim
Tangent line: | (1) = 4({ 1) | + 1 = 4{ 4 | = 4{ 5
25. (a) Using (4) with I ({) = 5{@(1 + {2 ) and the point (2> 2), we have
(b)
5(2 + k)
2
I
(2
+
k)
I
(2)
1
+
(2 + k)2
0
I (2) = lim
= lim
k0
k0
k
k
5k + 10
5k + 10 2(k2 + 4k + 5)
2
2
k2 + 4k + 5
= lim k + 4k + 5
= lim
k0
k0
k
k
2k2 3k
k(2k 3)
2k 3
3
= lim
= lim 2
=
k0 k(k2 + 4k + 5)
k0 k(k2 + 4k + 5)
k0 k + 4k + 5
5
= lim
So an equation of the tangent line at (2> 2) is | 2 = 35 ({ 2) or | = 35 { + 16
5 .
26. (a) Using (4) with J({) = 4{2 {3 , we have
J(d + k) J(d)
[4(d + k)2 (d + k)3 ] (4d2 d3 )
= lim
k0
k0
k
k
J0 (d) = lim
4d2 + 8dk + 4k2 (d3 + 3d2 k + 3dk2 + k3 ) 4d2 + d3
8dk + 4k2 3d2 k 3dk2 k3
= lim
k0
k0
k
k
= lim
k(8d + 4k 3d2 3dk k2 )
= lim (8d + 4k 3d2 3dk k2 ) = 8d 3d2
k0
k0
k
= lim
At the point (2> 8), J0 (2) = 16 12 = 4, and an equation of the
(b)
tangent line is | 8 = 4({ 2), or | = 4{. At the point (3> 9),
J0 (3) = 24 27 = 3, and an equation of the tangent line is
| 9 = 3({ 3), or | = 3{ + 18=
27. Use (4) with i ({) = 3{2 4{ + 1.
i (d + k) i(d)
[3(d + k)2 4(d + k) + 1] (3d2 4d + 1)]
= lim
k0
k0
k
k
i 0 (d) = lim
3d2 + 6dk + 3k2 4d 4k + 1 3d2 + 4d 1
6dk + 3k2 4k
= lim
k0
k0
k
k
= lim
= lim
k0
k(6d + 3k 4)
= lim (6d + 3k 4) = 6d 4
k0
k
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2 6 DERIVATIVES AND
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ยค
129
28. Use (4) with i (w) = 2w3 + w.
i (d + k) i(d)
[2(d + k)3 + (d + k)] (2d3 + d)
= lim
k0
k0
k
k
i 0 (d) = lim
2d3 + 6d2 k + 6dk2 + 2k3 + d + k 2d3 d
6d2 k + 6dk2 + 2k3 + k
= lim
k0
k0
k
k
= lim
k(6d2 + 6dk + 2k2 + 1)
= lim (6d2 + 6dk + 2k2 + 1) = 6d2 + 1
k0
k0
k
= lim
29. Use (4) with i (w) = (2w + 1)@(w + 3).
2(d + k) + 1 2d + 1
i (d + k) i (d)
(d + k) + 3
d+3
(2d + 2k + 1)(d + 3) (2d + 1)(d + k + 3)
i (d) = lim
= lim
= lim
k0
k0
k0
k
k
k(d + k + 3)(d + 3)
0
(2d2 + 6d + 2dk + 6k + d + 3) (2d2 + 2dk + 6d + d + k + 3)
k0
k(d + k + 3)(d + 3)
5k
5
5
= lim
=
= lim
k0 k(d + k + 3)(d + 3)
k0 (d + k + 3)(d + 3)
(d + 3)2
= lim
30. Use (4) with i ({) = {2 = 1@{2 .
1
1
d2 (d + k)2
2
2
i (d + k) i (d)
(d + k)
d
d2 (d + k)2
d2 (d2 + 2dk + k2 )
= lim
= lim
= lim
i 0 (d) = lim
k0
k0
k0
k0
k
k
k
kd2 (d + k)2
2dk k2
k(2d k)
2d k
2
2d
= lim
= lim 2
= 2 2 = 3
k0 kd2 (d + k)2
k0 kd2 (d + k)2
k0 d (d + k)2
d (d )
d
= lim
31. Use (4) with i ({) =
1 2{.
i (d + k) i (d)
i (d) = lim
= lim
k0
k0
k
0
s
1 2(d + k) 1 2d
k
2
s
2
s
s
1 2(d + k)
1 2d
1 2(d + k) 1 2d
1 2(d + k) + 1 2d
s
= lim
ยทs
= lim
k0
k
1 2(d + k) + 1 2d k0 k
1 2(d + k) + 1 2d
= lim
k0
(1 2d 2k) (1 2d)
2k
s
= lim s
k0
k
1 2(d + k) + 1 2d
k
1 2(d + k) + 1 2d
2
2
1
2
= lim s
=
=
=
k0
1 2d + 1 2d
2 1 2d
1 2d
1 2(d + k) + 1 2d
4
.
1{
32. Use (4) with i ({) =
4
4
s
1
d
1
(d
+
k)
i(d + k) i (d)
= lim
i 0 (d) = lim
k0
k0
k
k
1d 1dk
1d 1dk
1dk 1d
= 4 lim
= 4 lim
k0
k0 k 1 d k
k
1d
INSTRUCTOR USE ONLY
[continued]
[conti
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CHAPTER 2
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LIMITS AND
ND DERIVATIVES
VE
1d 1dk
1d+ 1dk
( 1 d)2 ( 1 d k)2
= 4 lim
ยท
= 4 lim
k0 k 1 d k
k0 k 1 d k
1d
1d+ 1dk
1 d( 1 d + 1 d k)
(1 d) (1 d k)
k
= 4 lim
k0 k 1 d k
k0 k 1 d k
1 d( 1 d + 1 d k)
1 d( 1 d + 1 d k)
= 4 lim
1
1
= 4 lim
=4ยท
k0
1 d 1 d( 1 d + 1 d)
1 d k 1 d( 1 d + 1 d k)
=
2
2
4
=
=
1
1@2
(1 d) (1 d)
(1 d)3@2
(1 d)(2 1 d)
Note that the answers to Exercises 33 โ 38 are not unique.
(1 + k)10 1
= i 0 (1), where i ({) = {10 and d = 1.
k0
k
(1 + k)10 1
Or: By (4), lim
= i 0 (0), where i ({) = (1 + {)10 and d = 0.
k0
k
4
16 + k 2
= i 0 (16), where i ({) = 4 { and d = 16.
34. By (4), lim
k0
k
4
16 + k 2
Or: By (4), lim
= i 0 (0), where i ({) = 4 16 + { and d = 0.
k0
k
33. By (4), lim
2{ 32
= i 0 (5), where i({) = 2{ and d = 5.
{5 { 5
35. By Equation 5, lim
36. By Equation 5, lim
{@4
tan { 1
= i 0 (@4), where i ({) = tan { and d = @4.
{ @4
cos( + k) + 1
= i 0 (), where i ({) = cos { and d = .
k
cos( + k) + 1
Or: By (4), lim
= i 0 (0), where i ({) = cos( + {) and d = 0.
k0
k
37. By (4), lim
k0
w4 + w 2
= i 0 (1), where i (w) = w4 + w and d = 1.
w1
w1
38. By Equation 5, lim
i (5 + k) i (5)
[100 + 50(5 + k) 4=9(5 + k)2 ] [100 + 50(5) 4=9(5)2 ]
= lim
k0
k0
k
k
2
(100 + 250 + 50k 4=9k 49k 122=5) (100 + 250 122=5)
4=9k2 + k
= lim
= lim
k0
k0
k
k
k(4=9k + 1)
= lim
= lim (4=9k + 1) = 1 m@s
k0
k0
k
39. y(5) = i 0 (5) = lim
The speed when w = 5 is |1| = 1 m@s.
i (5 + k) i (5)
[(5 + k)1 (5 + k)] (51 5)
= lim
k0
k0
k
k
40. y(5) = i 0 (5) = lim
5 5k(5 + k) (5 + k)
1
1
1
1
5k +5
k
5(5 + k)
5
+
k
5
5
+
k
5
= lim
= lim
= lim
k0
k0
k0
k
k
k
5 25k 5k2 5 k
5k2 26k
k(5k 26)
5k 26
26
= lim
= lim
= lim
=
m@s
k0
k0 5k(5 + k)
k0 5k(5 + k)
k0 5(5 + k)
5k(5 + k)
25
= lim
26
=
The speed when w = 5 is 26
25
25 = 1=04 m@s.
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SECTION
EC
22.66 DERIVATIVES AND
A RATES OF CHANGE
ยค
41. 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.
42. The slope of the tangent (that is, the rate of change of temperature with respect
to time) at w = 1 h seems to be about
43. (a) (i) [2002> 2006]:
75 168
132 0
0=7 F@min.
233 141
92
Q(2006) Q(2002)
=
=
= 23 millions of cell phone subscribers per year
2006 2002
4
4
(ii) [2002> 2004]:
Q(2004) Q(2002)
182 141
41
=
=
= 20=5 millions of cell phone subscribers per year
2004 2002
2
2
(iii) [2000> 2002]:
141 109
32
Q(2002) Q(2000)
=
=
= 16 millions of cell phone subscribers per year
2002 2000
2
2
(b) Using the values from (ii) and (iii), we have
20=5 + 16
= 18=25 millions of cell phone subscribers per year.
2
(c) Estimating A as (2000> 107) and B as (2004> 175), the slope at 2002
is
68
175 107
=
= 17 millions of cell phone subscribers per
2004 2000
4
year.
44. (a) (i) [2005> 2007]:
Q(2007) Q(2005)
15,011 10,241
4770
=
=
= 2385 locations per year
2007 2005
2
2
(ii) [2005> 2006]:
12,440 10,241
Q(2006) Q(2005)
=
= 2199 locations per year
2006 2005
1
(iii) [2004> 2005]:
10,241 8569
Q(2005) Q(2004)
=
= 1672 locations per year
2005 2004
1
(b) Using the values from (ii) and (iii), we have
2199 + 1672
= 1935= 5 locations per year.
2
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CHAPTER 2
LIMITS AND
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VE
(c) Estimating A as (2004> 8300) and B as (2006> 12,200), the slope at
2005 is
45. (a) (i)
3900
12,200 8300
=
= 1950 locations per year.
2006 2004
2
F(105) F(100)
6601=25 6500
F
=
=
= $20=25@unit.
{
105 100
5
F(101) F(100)
6520=05 6500
F
=
=
= $20=05@unit.
{
101 100
1
5000 + 10(100 + k) + 0=05(100 + k)2 6500
F(100 + k) F(100)
20k + 0=05k2
(b)
=
=
k
k
k
= 20 + 0=05k, k 6= 0
(ii)
F(100 + k) F(100)
= lim (20 + 0=05k) = $20@unit.
k0
k0
k
2
2
w+k
w
46. Y = Y (w + k) Y (w) = 100,000 1
100,000 1
60
60
w+k
(w + k)2
w
w2
k
2wk
k2
= 100,000 1
+
1
+
= 100,000 +
+
30
3600
30
3600
30
3600
3600
So the instantaneous rate of change is lim
=
100,000
250
k (120 + 2w + k) =
k (120 + 2w + k)
3600
9
Dividing Y by k and then letting k
(w 60) gal@min.
0, we see that the instantaneous rate of change is 500
9
w
Flow rate (gal@min)
Water remaining Y (w) (gal)
0
3333=3
100> 000
10
2777=7
69> 444=4
20
2222=2
44> 444=4
30
1666=6
25> 000
40
1111=1
11> 111=1
50
555=5
2> 777=7
60
0
0
The magnitude of the ow rate is greatest at the beginning and gradually decreases to 0.
47. (a) i 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 i 0 ({) will decrease because more efcient use is made of start-up costs as { increases. But
eventually i 0 (({)) might increase due to large-scale operations.
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22.66 DERIVATIVES AND
A RATES OF CHANGE
ยค
133
48. (a) i 0 (5) is the rate of growth of the bacteria population when w = 5 hours. Its units are bacteria per hour.
(b) With unlimited space and nutrients, i 0 should increase as w increases; so i 0 (5) ? i 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.
49. W 0 (10) is the rate at which the temperature is changing at 10:00 AM . To estimate the value of W 0 (10), we will average the
65 76
W (8) W (10)
=
= 5=5 and
8 10
2
W (12) W (10)
W (w) W (10)
85 76
D+E
5=5 + 4=5
E=
=
= 4=5. Then W 0 (10) = lim
=
= 5 F@ h.
w10
12 10
2
w 10
2
2
difference quotients obtained using the times w = 8 and w = 12. Let D =
50. (a) i 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 i 0 (8) are pounds@(dollars@pound).
(b) i 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.
51. (a) V 0 (W ) is the rate at which the oxygen solubility changes with respect to the water temperature. Its units are (mg@L)@ C.
(b) For W = 16 C, it appears that the tangent line to the curve goes through the points (0> 14) and (32> 6). So
V 0 (16)
6 14
8
=
= 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.
52. (a) V 0 (W ) 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 W = 15 C, it appears the tangent line to the curve goes through the points (10> 25) and (20> 32). So
V 0 (15)
32 25
= 0=7 (cm@s)@ C. This tells us that at W = 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 W = 25 C, we can use the points (20> 35) and (25> 25) to
obtain V 0 (25)
25 35
= 2 (cm@s)@ C. As it gets warmer than 20 C, the maximum sustainable speed decreases
25 20
rapidly.
53. Since i ({) = { sin(1@{) when { 6= 0 and i (0) = 0, we have
i (0 + k) i (0)
k sin(1@k) 0
= lim
= lim sin(1@k). This limit does not exist since sin(1@k) takes the
k0
k0
k
k
values 1 and 1 on any interval containing 0. (Compare with Example 4 in Section 2.2.)
i 0 (0) = lim
k0
54. Since i({) = {2 sin(1@{) when { 6= 0 and i (0) = 0, we have
i (0 + k) i (0)
k2 sin(1@k) 0
1
= lim
= lim k sin(1@k). Since 1 sin 1, we have
k0
k0
k0
k
k
k
i 0 (0) = lim
1
1
|k| |k| k sin |k|. Because lim ( |k|) = 0 and lim |k| = 0, we know that
k0
k0
k
k
1
= 0 by the Squeeze Theorem. Thus, i 0 (0) = 0.
lim k sin
k0
k
|k| |k| sin
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LIMITS AND
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VE
2.7 The Derivative as a Function
1. It appears that i is an odd function, so i 0 will be an even
functionโthat is, i 0 (d) = i 0 (d).
(a) i 0 (3)
0=2
(b) i 0 (2)
0
(c) i 0 (1)
1
(d) i 0 (0)
2
(e) i 0 (1)
1
(f ) i 0 (2)
0
(g) i 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 i ,
it appears that i 0 (0)
(b) i 0 (1)
0
(c) i 0 (2)
1=5
(d) i 0 (3)
1=3
(e) i 0 (4)
0=8
(f ) i 0 (5)
0=3
(g) i 0 (6)
0
(h) i 0 (7)
0=2
6.
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)0 = IV, since from left to right, the slopes of the tangents to graph (b) start out at a xed positive quantity, then suddenly
become negative, then positive again. The discontinuities in graph IV indicate sudden changes in the slopes of the tangents.
(c)0 = I, since the slopes of the tangents to graph (c) are negative for { ? 0 and positive for { A 0, as are the function values of
graph I.
(d)0 = III, since from left to right, the slopes of the tangents to graph (d) are positive, then 0, then negative, then 0, then
positive, then 0, then negative again, and the function values in graph III follow the same pattern.
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THE DERIVATIVE
DERIV
AS A FUNCTION
ยค
135
Hints for Exercises 4 โ11: First plot {-intercepts on the graph of i 0 for any horizontal tangents on the graph of i . Look for any corners on the graph
of i โ there will be a discontinuity on the graph of i 0 . On any interval where i has a tangent with positive (or negative) slope, the graph of i 0 will be
positive (or negative). If the graph of the function is linear, the graph of i 0 will be a horizontal line.
4.
5.
6.
7.
8.
9.
10.
11.
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LIMITS AND
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VE
12. The slopes of the tangent lines on the graph of | = S (w) are always
positive, so the |-values of | = S 0(w) are always positive. These values start
out relatively small and keep increasing, reaching a maximum at about
w = 6. Then the |-values of | = S 0(w) decrease and get close to zero. The
graph of S 0 tells us that the yeast culture grows most rapidly after 6 hours
and then the growth rate declines.
13. It appears that there are horizontal tangents on the graph of P for w = 1963
and w = 1971. Thus, there are zeros for those values of w on the graph of
P 0 . The derivative is negative for the years 1963 to 1971.
14. See Figure 1 in Section 3.3.
15.
16.
The slope at 0 appears to be 1 and the slope at 1
As { increases toward 1, i 0 ({) decreases from very large
appears to be 2=7. As { decreases, the slope gets
numbers to 1. As { becomes large, i 0 ({) gets closer to 0.
closer to 0. Since the graphs are so similar, we might
As a guess, i 0 ({) = 1@{2 or i 0 ({) = 1@{ makes sense.
guess that i 0 ({) = h{ .
17. (a) By zooming in, we estimate that i 0 (0) = 0, i 0 12 = 1, i 0 (1) = 2,
and i 0 (2) = 4.
(b) By symmetry, i 0 ({) = i 0 ({). So i 0 12 = 1, i 0 (1) = 2,
and i 0 (2) = 4.
(c) It appears that i 0 ({) is twice the value of {, so we guess that i 0 ({) = 2{.
i ({ + k) i ({)
({ + k)2 {2
= lim
k0
k0
k
k
2
{ + 2k{ + k2 {2
2k{ + k2
k(2{ + k)
= lim
= lim
= lim (2{ + k) = 2{
= lim
k0
k0
k0
k0
k
k
k
(d) i 0 ({) = lim
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SECTION 22.77
18. (a) By zooming in, we estimate that i 0 (0) = 0, i 0 12
i 0 (1)
3, i 0 (2)
12, and i 0 (3)
0=75,
THE DERIVATIVE
DERIV
AS A FUNCTION
ยค
(b) By symmetry, i 0 ({) = i 0 ({). So i 0 12
i 0 (1)
27.
3, i 0 (2)
12, and i 0 (3)
137
0=75,
27.
(d) Since i 0 (0) = 0, it appears that i 0 may have the
(c)
form i 0 ({) = d{2 . Using i 0 (1) = 3, we have d = 3,
so i 0 ({) = 3{2 .
i({ + k) i ({)
({ + k)3 {3
({3 + 3{2 k + 3{k2 + k3 ) {3
= lim
= lim
k0
k0
k0
k
k
k
2
2
2
2
3
3{ k + 3{k + k
k(3{ + 3{k + k )
= lim
= lim
= lim (3{2 + 3{k + k2 ) = 3{2
k0
k0
k0
k
k
(e) i 0 ({) = lim
i ({ + k) i({)
19. i ({) = lim
= lim
k0
k0
k
0
1
1
2 ({ + k) 3
k
1
1
2{ 3
1
= lim 2
k0
{ + 12 k 13 12 { + 13
k
1
k
= lim 2 = lim 12 = 12
k0 k
k0
Domain of i = domain of i 0 = R.
20. i 0 ({) = lim
k0
= lim
k0
i({ + k) i ({)
[p({ + k) + e] (p{ + e)
p{ + pk + e p{ e
= lim
= lim
k0
k0
k
k
k
pk
= lim p = p
k0
k
Domain of i = domain of i 0 = R.
5(w + k) 9(w + k)2 (5w 9w2 )
i (w + k) i(w)
= lim
21. i (w) = lim
k0
k0
k
k
0
5w + 5k 9(w2 + 2wk + k2 ) 5w + 9w2
5w + 5k 9w2 18wk 9k2 5w + 9w2
= lim
k0
k0
k
k
= lim
5k 18wk 9k2
k(5 18w 9k)
= lim
= lim (5 18w 9k) = 5 18w
k0
k0
k0
k
k
= lim
Domain of i = domain of i 0 = R.
1=5({ + k)2 ({ + k) + 3=7 1=5{2 { + 3=7
i({ + k) i ({)
= lim
k0
k0
k
k
22. i 0 ({) = lim
1=5{2 + 3{k + 1=5k2 { k + 3=7 1=5{2 + { 3=7
3{k + 1=5k2 k
= lim
k0
k0
k
k
= lim
= lim (3{ + 1=5k 1) = 3{ 1
k0
Domain of i = domain of i 0 = R.
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LIMITS AND
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VE
i({ + k) i ({)
[({ + k)2 2({ + k)3 ] ({2 2{3 )
= lim
k0
k0
k
k
23. i 0 ({) = lim
{2 + 2{k + k2 2{3 6{2 k 6{k2 2k3 {2 + 2{3
k0
k
= lim
2{k + k2 6{2 k 6{k2 2k3
k(2{ + k 6{2 6{k 2k2 )
= lim
k0
k0
k
k
= lim (2{ + k 6{2 6{k 2k2 ) = 2{ 6{2
= lim
k0
Domain of i = domain of i 0 = R.
{ + k + { + k ({ + { )
i ({ + k) i ({)
= lim
k0
k0
k
k
%
&
({ + k) {
{+k {
{+k+ {
k
= lim 1 +
+
ยท
= lim
k0
k0
k
k
{+k+ {
k {+k+ {
1
1
1
=1+
=1+
= lim 1 +
k0
{
+
{
2
{
{+k+ {
24. i 0({) = lim
Domain of i = [0> ), domain of i 0 = (0> ).
j({ + k) j({)
= lim
25. j ({) = lim
k0
k0
k
0
= lim
k0
%s
&
s
1 + 2({ + k) 1 + 2{
1 + 2({ + k) + 1 + 2{
s
k
1 + 2({ + k) + 1 + 2{
(1 + 2{ + 2k) (1 + 2{)
1
2
2
ks
l = lim
=
=
k0
2
1
+
2{
1
+
2{
1 + 2{ + 2k + 1 + 2{
k
1 + 2({ + k) + 1 + 2{
Domain of j = 12 > , domain of j0 = 12 > .
({ + k)2 1
{2 1
i({ + k) i ({)
2({ + k) 3
2{ 3
= lim
26. i 0 ({) = lim
k0
k0
k
k
[({ + k)2 1](2{ 3) [2({ + k) 3]({2 1)
[2({ + k) 3](2{ 3)
= lim
k0
k
({2 + 2{k + k2 1)(2{ 3) (2{ + 2k 3)({2 1)
k0
k[2({ + k) 3](2{ 3)
= lim
(2{3 + 4{2 k + 2{k2 2{ 3{2 6{k 3k2 + 3) (2{3 + 2{2 k 3{2 2{ 2k + 3)
k0
k(2{ + 2k 3)(2{ 3)
= lim
4{2 k + 2{k2 6{k 3k2 2{2 k + 2k
k(2{2 + 2{k 6{ 3k + 2)
= lim
k0
k0
k(2{ + 2k 3)(2{ 3)
k(2{ + 2k 3)(2{ 3)
= lim
2{2 + 2{k 6{ 3k + 2
2{2 6{ + 2
=
k0 (2{ + 2k 3)(2{ 3)
(2{ 3)2
= lim
Domain of i = domain of i 0 = (> 32 )
( 32 > ).
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SECTION 22.77
THE DERIVATIVE
DERIV
AS A FUNCTION
ยค
4(w + k)
4(w + k)(w + 1) 4w(w + k + 1)
4w
J(w
+
k)
J(w)
(w
+
k)
+
1
w
+
1
(w + k + 1)(w + 1)
= lim
= lim
27. J0(w) = lim
k0
k0
k0
k
k
k
2
2
4w + 4kw + 4w + 4k 4w + 4kw + 4w
4k
= lim
= lim
k0
k0 k(w + k + 1)(w + 1)
k(w + k + 1)(w + 1)
4
= lim
k0 (w + k + 1)(w + 1)
=
4
(w + 1)2
Domain of J = domain of J0 = (> 1)
(1> ).
w w+k
1
1
j(w + k) j(w)
w w+k
w+ w+k
w+k
w+k w
w
0
ยท
28. j (w) = lim
= lim
= lim
= lim
k0
k0
k0
k0
k
k
k
k w+k w
w+ w+k
= lim
k0 k
w (w + k)
k
1
= lim
= lim
k0
k0
w+k w w+ w+k
k w+k w w+ w+k
w+k w w+ w+k
1
1
1
= = 3@2
=
2w
w w w+ w
w 2 w
Domain of j = domain of j0 = (0> ).
4
{ + 4{3 k + 6{2 k2 + 4{k3 + k4 {4
i({ + k) i ({)
({ + k)4 {4
29. i ({) = lim
= lim
= lim
k0
k0
k0
k
k
k
3
2 2
3
4
3
4{ k + 6{ k + 4{k + k
= lim
= lim 4{ + 6{2 k + 4{k2 + k3 = 4{3
k0
k0
k
0
Domain of i = domain of i 0 = R.
({ + k)2 + 1
{2 + 1
i({ + k) i ({)
[({ + k) + 1@({ + k)] ({ + 1@{)
{+k
{
= lim
= lim
30. (a) i 0 ({) = lim
k0
k0
k0
k
k
k
{[({ + k)2 + 1] ({ + k)({2 + 1)
({3 + 2k{2 + {k2 + {) ({3 + { + k{2 + k)
= lim
k0
k0
k({ + k){
k({ + k){
= lim
k{2 + {k2 k
k({2 + {k 1)
{2 + {k 1
{2 1
1
= lim
= lim
=
, or 1 2
k0
k0
k0
k({ + k){
k({ + k){
({ + k){
{2
{
= lim
(b) Notice that i 0 ({) = 0 when i has a horizontal tangent, i 0 ({) is
positive when the tangents have positive slope, and i 0 ({) is
negative when the tangents have negative slope. Both functions
are discontinuous at { = 0.
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CHAPTER 2
LIMITS AND
ND DERIVATIVES
VE
i ({ + k) i ({)
[({ + k)4 + 2({ + k)] ({4 + 2{)
= lim
k0
k0
k
k
31. (a) i 0 ({) = lim
{4 + 4{3 k + 6{2 k2 + 4{k3 + k4 + 2{ + 2k {4 2{
k0
k
= lim
4{3 k + 6{2 k2 + 4{k3 + k4 + 2k
k(4{3 + 6{2 k + 4{k2 + k3 + 2)
= lim
k0
k0
k
k
= lim
= lim (4{3 + 6{2 k + 4{k2 + k3 + 2) = 4{3 + 2
k0
(b) Notice that i 0 ({) = 0 when i has a horizontal tangent, i 0 ({) is
positive when the tangents have positive slope, and i 0 ({) is
negative when the tangents have negative slope.
(w + k)2 w + k w2 w
i (w + k) i (w)
= lim
32. (a) i (w) = lim
k0
k0
k
k
2
2
2
w + 2kw + k w + k w + w
w w+k
2kw + k2
= lim
= lim
+
k0
k0
k
k
k
w (w + k)
w w+k
w+ w+k
k(2w + k)
= lim 2w + k +
+
ยท
= lim
k0
k0
k
k
w+ w+k
k( w + w + k )
k
1
1
= lim 2w + k +
= lim 2w + k +
= 2w
k0
k0
k( w + w + k )
w+ w+k
2 w
0
(b) Notice that i 0 (w) = 0 when i has a horizontal tangent, i 0 (w) is
positive when the tangents have positive slope, and i 0 (w) is
negative when the tangents have negative slope.
33. (a) X 0 (w) is the rate at which the unemployment rate is changing with respect to time. Its units are percent per year.
(b) To nd X 0 (w), we use lim
k0
X (w + k) X(w)
k
X(w + k) X (w)
for small values of k.
k
X (1999) X(1998)
4=2 4=5
=
= 0=30
1999 1998
1
For 1998: X 0 (1998)
For 1999: We estimate X 0 (1999) by using k = 1 and k = 1, and then average the two results to obtain a nal estimate.
k = 1 X 0 (1999)
k = 1 X 0 (1999)
4=5 4=2
X (1998) X (1999)
=
= 0=30;
1998 1999
1
4=0 4=2
X (2000) X (1999)
=
= 0=20.
2000 1999
1
So we estimate that X 0 (1999)
1
2 [(0=30) + (0=20)] = 0=25.
w
1998
1999
2000
2001
2002
2003
2004
2005
2006
2007
0
0=30
0=25
0=25
0=90
0=65
0=15
0=45
0=45
0=25
0=00
X (w)
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141
34. (a) S 0 (w) 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).
S (w + k) S (w)
k0
k
(b) To nd S 0 (w), we use lim
For 1950: S 0 (1950)
S (w + k) S (w)
for small values of k.
k
35=7 31=1
S (1960) S (1950)
=
= 0=46
1960 1950
10
For 1960: We estimate S 0 (1960) by using k = 10 and k = 10, and then average the two results to obtain a
nal estimate.
k = 10 S 0 (1960)
31=1 35=7
S (1950) S (1960)
=
= 0=46
1950 1960
10
k = 10 S 0 (1960)
S (1970) S (1960)
34=0 35=7
=
= 0=17
1970 1960
10
So we estimate that S 0 (1960)
1
[0=46 + (0=17)] = 0=145.
2
w
0
S (w)
1950
1960
1970
1980
1990
2000
0=460
0=145
0=385
0=415
0=115
0=000
(c)
(d) We could get more accurate values for S 0 (w) by obtaining data for the mid-decade years 1955, 1965, 1975, 1985, and 1995.
35. i is not differentiable at { = 4, because the graph has a corner there, and at { = 0, because there is a discontinuity there.
36. i is not differentiable at { = 0, because there is a discontinuity there, and at { = 3, because the graph has a vertical tangent
there.
37. i is not differentiable at { = 1, because the graph has a vertical tangent there, and at { = 4, because the graph has a corner
there.
38. i is not differentiable at { = 1, because there is a discontinuity there, and at { = 2, because the graph has a corner there.
39. As we zoom in toward (1> 0), the curve appears more and more like a straight
line, so i ({) = { +
s
|{| is differentiable at { = 1. But no matter how much
we zoom in toward the origin, the curve doesnโt straighten outโwe canโt
eliminate the sharp point (a cusp). So i is not differentiable at { = 0.
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VE
40. As we zoom in toward (0> 1), the curve appears more and more like a straight
line, so i is differentiable at { = 0. But no matter how much we zoom in toward
(1> 0) or (1> 0), the curve doesnโt straighten outโwe canโt eliminate the sharp
point (a cusp). So i is not differentiable at { = ยฑ1.
41. d = i , e = i 0 , f = i 00 . We can see this because where d has a horizontal tangent, e = 0, and where e has a horizontal tangent,
f = 0. We can immediately see that f can be neither i nor i 0 , since at the points where f has a horizontal tangent, neither d
nor e is equal to 0.
42. Where g has horizontal tangents, only f is 0, so g0 = f. f has negative tangents for { ? 0 and e is the only graph that is
negative for { ? 0, so f0 = e. e has positive tangents on R (except at { = 0), and the only graph that is positive on the same
domain is d, so e0 = d. We conclude that g = i , f = i 0 , e = i 00 , and d = i 000 .
43. We can immediately see that d is the graph of the acceleration function, since at the points where d has a horizontal tangent,
neither f nor e is equal to 0. Next, we note that d = 0 at the point where e has a horizontal tangent, so e must be the graph of
the velocity function, and hence, e0 = d. We conclude that f is the graph of the position function.
44. d must be the jerk since none of the graphs are 0 at its high and low points. d is 0 where e has a maximum, so e0 = d. e is 0
where f has a maximum, so f0 = e. We conclude that g is the position function, f is the velocity, e is the acceleration, and d is
the jerk.
i({ + k) i ({)
[3({ + k)2 + 2({ + k) + 1] (3{2 + 2{ + 1)
= lim
k0
k0
k
k
45. i 0 ({) = lim
(3{2 + 6{k + 3k2 + 2{ + 2k + 1) (3{2 + 2{ + 1)
6{k + 3k2 + 2k
= lim
k0
k0
k
k
= lim
= lim
k0
k(6{ + 3k + 2)
= lim (6{ + 3k + 2) = 6{ + 2
k0
k
i 0 ({ + k) i 0 ({)
[6({ + k) + 2] (6{ + 2)
(6{ + 6k + 2) (6{ + 2)
= lim
= lim
k0
k0
k0
k
k
k
i 00 ({) = lim
= lim
k0
6k
= lim 6 = 6
k0
k
We see from the graph that our answers are reasonable because the graph of
i 0 is that of a linear function and the graph of i 00 is that of a constant
function.
i({ + k) i ({)
[({ + k)3 3({ + k)] ({3 3{)
= lim
k0
k0
k
k
3
2
2
3
({ + 3{ k + 3{k + k 3{ 3k) ({3 3{)
3{2 k + 3{k2 + k3 3k
= lim
= lim
k0
k0
k
k
46. i 0 ({) = lim
k(3{2 + 3{k + k2 3)
= lim (3
(3{
{2 + 3{k
3{k + k2 3) = 3{
3{2 3
k0
k
0
k0
k
0
k
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THE DERIVATIVE
DERIV
AS A FUNCTION
ยค
i 0 ({ + k) i 0 ({)
[3({ + k)2 3] (3{2 3)
(3{2 + 6{k + 3k2 3) (3{2 3)
= lim
= lim
k0
k0
k0
k
k
k
i 00 ({) = lim
6{k + 3k2
k(6{ + 3k)
= lim
= lim (6{ + 3k) = 6{
k0
k0
k0
k
k
= lim
We see from the graph that our answers are reasonable because the graph of
i 0 is that of a even function (i is an odd function) and the graph of i 00 is
that of an odd function. Furthermore, i 0 = 0 when i has a horizontal
tangent and i 00 = 0 when i 0 has a horizontal tangent.
2({ + k)2 ({ + k)3 (2{2 {3 )
i ({ + k) i ({)
= lim
47. i ({) = lim
k0
k0
k
k
0
k(4{ + 2k 3{2 3{k k2 )
= lim (4{ + 2k 3{2 3{k k2 ) = 4{ 3{2
k0
k0
k
= lim
4({ + k) 3({ + k)2 (4{ 3{2 )
i 0 ({ + k) i 0 ({)
k(4 6{ 3k)
= lim
= lim
k0
k0
k0
k
k
k
i 00 ({) = lim
= lim (4 6{ 3k) = 4 6{
k0
i 00 ({ + k) i 00 ({)
[4 6({ + k)] (4 6{)
6k
= lim
= lim
= lim (6) = 6
k0
k0
k0 k
k0
k
k
i 000 ({) = lim
i 000 ({ + k) i 000 ({)
6 (6)
0
= lim
= lim = lim (0) = 0
k0
k0
k0
k
k
k k0
i (4) ({) = lim
The graphs are consistent with the geometric interpretations of the
derivatives because i 0 has zeros where i has a local minimum and a local
maximum, i 00 has a zero where i 0 has a local maximum, and i 000 is a
constant function equal to the slope of i 00 .
48. (a) Since we estimate the velocity to be a maximum at w = 10, the acceleration is 0 at w = 10.
(b) Drawing a tangent line at w = 10 on the graph of d, d appears to decrease by 10 ft@s2 over a period of 20 s.
So at w = 10 s, the jerk is approximately 10@20 = 0=5 (ft@s2 )@s or ft@s3 .
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LIMITS AND
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VE
49. (a) Note that we have factored { d as the difference of two cubes in the third step.
i ({) i (d)
{1@3 d1@3
{1@3 d1@3
= lim
= lim 1@3
{d
{d
{d ({
{d
{d
d1@3 )({2@3 + {1@3 d1@3 + d2@3 )
i 0(d) = lim
1
= lim
{d {2@3 + {1@3 d1@3 + d2@3
=
1
or 13 d2@3
3d2@3
3
i(0 + k) i(0)
k0
1
(b) i (0) = lim
= lim
= lim 2@3 . This function increases without bound, so the limit does not
k0
k0
k0 k
k
k
0
exist, and therefore i 0(0) does not exist.
(c) lim |i 0 ({)| = lim
1
{0 3{2@3
{0
= and i is continuous at { = 0 (root function), so i has a vertical tangent at { = 0.
j({) j(0)
{2@3 0
1
= lim
= lim 1@3 , which does not exist.
{0
{0
{0 {
{0
{
50. (a) j 0(0) = lim
j({) j(d)
({1@3 d1@3 )({1@3 + d1@3 )
{2@3 d2@3
= lim
= lim 1@3
{d
{d
{d ({
{d
{d
d1@3 )({2@3 + {1@3 d1@3 + d2@3 )
(b) j 0(d) = lim
{1@3 + d1@3
= lim
{d {2@3 + {1@3 d1@3 + d2@3
=
2d1@3
2
= 1@3 or 23 d1@3
3d2@3
3d
(c) j({) = {2@3 is continuous at { = 0 and
lim |j0({)| = lim
2
{0 3 |{|1@3
{0
(d)
= . This shows that
j has a vertical tangent line at { = 0.
51. i ({) = |{ 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
i ({) i (6)
|{ 6| 0
{6
= lim
= lim
= lim 1 = 1, and the left-hand limit
+
+
{6
{6
{6
{6 { 6
{6+
i({) i(6)
|{ 6| 0
6{
= lim
= lim
= lim (1) = 1. Since these limits are not equal,
{6
{6
{6
{6 { 6
{6
i ({) i (6)
does not exist and i is not differentiable at 6.
{6
+
1
if { A 6
0
0
However, a formula for i is i ({) =
1 if { ? 6
i 0 (6) = lim
{6
Another way of writing the formula is i 0 ({) =
{6
.
|{ 6|
52. i ({) = [[{]] is not continuous at any integer q, so i is not differentiable
at q by the contrapositive of Theorem 4. If d is not an integer, then i is
constant on an open interval containing d, so i 0(d) = 0. Thus,
i 0({) = 0, { not an integer.
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145
53. (a) If i is even, then
i 0 ({) = lim
k0
= lim
k0
i ({ + k) i ({)
i [({ k)] i ({)
= lim
k0
k
k
i ({ k) i({)
i ({ k) i ({)
= lim
k0
k
k
= lim
{0
[let { = k]
i ({ + {) i ({)
= i 0 ({)
{
Therefore, i 0 is odd.
(b) If i is odd, then
i 0 ({) = lim
k0
= lim
k0
i ({ + k) i ({)
i [({ k)] i ({)
= lim
k0
k
k
i({ k) + i ({)
i ({ k) i({)
= lim
k0
k
k
[let { = k]
i ({ + {) i ({)
= i 0 ({)
{0
{
= lim
Therefore, i 0 is even.
54. (a)
(b) The initial temperature of the water is close to room temperature because of the water that was in the pipes. When the
water from the hot water tank starts coming out, gW @gw is large and positive as W increases to the temperature of the water
in the tank. In the next phase, gW @gw = 0 as the water comes out at a constant, high temperature. After some time, gW @gw
becomes small and negative as the contents of the hot water tank are exhausted. Finally, when the hot water has run out,
gW @gw is once again 0 as the water maintains its (cold) temperature.
(c)
55.
In the right triangle in the diagram, let | be the side opposite angle ! and {
the side adjacent angle !. Then the slope of the tangent line c
is p = |@{ = tan !. Note that 0 ? ! ? 2 . We know (see Exercise 17)
that the derivative of i({) = {2 is i 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, ! = tan1 2
63 .
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2.8 What Does i 0 Say about i ?
1. (a) Since i 0 ({) ? 0 on (1> 4), i is decreasing on this interval. Since i 0 ({) A 0 on (0> 1) and (4> 5), i is increasing on these
intervals.
(b) At { = 1, i 0 ({) = 0 and i 0 changes from positive to negative there, i changes from increasing to decreasing and has a
local maximum at { = 1. At { = 4, i 0 ({) = 0 and i 0 changes from negative to positive there, i changes from decreasing
to increasing and has local minimum at { = 4.
(c) Since i(0) = 0, start at the origin. Draw an increasing function on (0> 1)
with a local maximum at { = 1. Now draw a decreasing function on (1> 4)
and the steepest slope should occur at { = 2=5 since thatโs where the
smallest value of i 0 occurs. Last, draw an increasing function on (4> 5)
making sure you have a local minimum at { = 4.
2. (a) i 0 ({) A 0 and i is increasing on (2> 0) and (2> 3). i 0 ({) ? 0 and i is
(c)
decreasing on (3> 2) and (0> 2).
(b) At { = 0, i 0 ({) = 0 and i 0 changes from positive to negative, so i has a
local maximum at { = 0. At { = 2 and { = 2, i 0 ({) = 0 and i 0 changes
from negative to positive, so i has local minima at { = 2 and { = 2.
3. (a) i 0 ({) A 0 and i is increasing on (2> 1), (0> 1), and (2> 3). i 0 ({) ? 0
(c)
and i is decreasing on (1> 0) and (1> 2).
(b) At { = 1 and { = 1, i 0 ({) = 0 and i 0 changes from positive to negative,
so i has local maxima at { = 1 and { = 1. At { = 0 and { = 2,
i 0 ({) = 0 and i 0 changes from negative to positive, so i has local minima
at { = 0 and { = 2.
4. (a) i 0 ({) A 0 and i is increasing on (2> 1) and (0> 1). i 0 ({) ? 0 and i is
(c)
decreasing on (1> 0) and (1> 2).
(b) At { = 1 and { = 1, i 0 ({) = 0 and i 0 changes from positive to negative,
so i has local maxima at { = 1 and { = 1. At { = 0, i 0 ({) = 0 and i 0
changes from negative to positive, so i has a local minimum at { = 0.
(The points at { = 2 and { = 2 are not part of the graph.)
5. The derivative i 0 is increasing when the slopes of the tangent lines of i are becoming larger as { increases. This seems to be
the case on the interval (2> 5). The derivative is decreasing when the slopes of the tangent lines of i are becoming smaller as {
increases, and this seems to be the case on (> 2) and (5> ). So i 0 is increasing on (2> 5) and decreasing on (> 2)
and (5> ).
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147
6. Call the curve with the smallest positive {-intercept j and the other curve k. Notice that where j is positive in the rst
quadrant, k is increasing. Thus, k = i and j = i 0 . Now i 0 (1) is positive since i 0 is above the {-axis there and i 00 (1)
appears to be zero since i has an inection point at { = 1. Therefore, i 00 (1) is greater than i 0 (1).
7. Call the curve with the positive |-intercept j and the other curve k. Notice that j has a maximum (horizontal tangent) at
{ = 0, but k 6= 0, so k cannot be the derivative of j. Also notice that where j is positive, k is increasing. Thus, k = i and
j = i 0 . Now i 0 (1) is negative since i 0 is below the {-axis there and i 00 (1) is positive since i is concave upward at { = 1.
Therefore, i 00 (1) is greater than i 0 (1).
8. (a)
(b)
(c) In part (a), the graph of | = h{ is a curve whose
slope is always positive and increasing. In
part (b), the graph of | = ln { is a curve whose
slope is always positive and decreasing.
9. If G(w) is the size of the decit as a function of time, then at the time of the speech G0 (w) A 0, but G00 (w) ? 0 because
G00 (w) = (G0 )0(w) is the rate of change of G0 (w).
10. (a) The rate of increase of the population is initially very small, then gets larger until it reaches a maximum at about
w = 8 hours, and decreases toward 0 as the population begins to level off.
(b) The rate of increase has its maximum value at w = 8 hours.
(c) The population function is concave upward on (0> 8) and concave downward
on (8> 18).
(d) At w = 8, the population is about 350, so the inection point is about (8> 350).
11. (a) The rate of increase of the population is initially very small, then
increases rapidly until about 1932 when it starts decreasing. The
rate becomes negative by 1936, peaks in magnitude in 1937, and
approaches 0 in 1940.
(b) Inection points (IP) appear to be at (1932> 2=5) and (1937> 4=3).
The rate of change of population density starts to decrease in 1932
and starts to increase in 1937. The rates of population increase and
decrease have their maximum values at those points.
12. (a) If the position function is increasing, then the particle is moving toward the right. This occurs on w-intervals (0> 2) and
(4> 6). If the function is decreasing, then the particle is moving toward the left โ that is, on (2> 4).
(b) The acceleration is the second derivative and is positive where the curve is concave upward. This occurs on (3> 6). The
acceleration is negative where the curve is concave downward โ that is, on (0> 3).
13. Most students learn more in the third hour of studying than in the eighth hour, so N(3) N(2) is larger than N(8) N(7).
In other words, as you begin studying for a test, the rate of knowledge gain is large and then starts to taper off, so N 0 (w)
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14. At rst the depth increases slowly because the base of the mug is wide.
But as the mug narrows, the coffee rises more quickly. Thus, the depth
g increases at an increasing rate and its graph is concave upward. The
rate of increase of g has a maximum where the mug is narrowest; that is,
when the mug is half full. It is there that the inection point (IP) occurs.
Then the rate of increase of g starts to decrease as the mug widens and
the graph becomes concave down.
15. (a) i is increasing where i 0 is positive, that is, on (0> 2), (4> 6), and (8> ); and decreasing where i 0 is negative, that is, on
(2> 4) and (6> 8).
(b) i has local maxima where i 0 changes from positive to negative, at { = 2 and at { = 6, and local minima where i 0 changes
from negative to positive, at { = 4 and at { = 8.
(c) i is concave upward (CU) where i 0 is increasing, that is, on (3> 6) and (6> ), and concave downward (CD) where i 0 is
decreasing, that is, on (0> 3).
(d) There is a point of inection where i changes from
(e)
being CD to being CU, that is, at { = 3.
16. (a) i is increasing where i 0 is positive, on (1> 6) and (8> ), and decreasing where i 0 is negative, on (0> 1) and (6> 8).
(b) i has a local maximum where i 0 changes from positive to negative, at { = 6, and local minima where i 0 changes from
negative to positive, at { = 1 and at { = 8.
(c) i is concave upward where i 0 is increasing, that is, on (0> 2), (3> 5), and (7> ), and concave downward where i 0 is
decreasing, that is, on (2> 3) and (5> 7).
(d) There are points of inection where i changes its
(e)
direction of concavity, at { = 2, { = 3, { = 5 and
{ = 7.
17. The function must be always decreasing (since the rst
derivative is always negative) and concave downward
18. The function must be always decreasing and concave
upward.
(since the second derivative is always negative).
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DOES i 0 SAY ABOUT i ?
19. i 0 (0) = i 0 (4) = 0
ยค
horizontal tangents at { = 0> 4.
i 0 ({) A 0 if { ? 0 i is increasing on (> 0).
i 0 ({) ? 0 if 0 ? { ? 4 or if { A 4 i is decreasing on (0> 4) and (4> ).
i 00 ({) A 0 if 2 ? { ? 4 i is concave upward on (2> 4).
i 00 ({) ? 0 if { ? 2 or { A 4 i is concave downward on (> 2)
and (4> ). There are inection points when { = 2 and 4.
20. i 0 ({) A 0 for all { 6= 1 with vertical asymptote { = 1, so i is increasing on
(> 1) and (1> ). i 00 ({) A 0 if { ? 1 or { A 3, and i 00 ({) ? 0 if 1 ? { ? 3,
so i is concave upward on (> 1) and (3> ), and concave downward on (1> 3).
There is an inection point when { = 3.
21. i 0 (0) = i 0 (2) = i 0 (4) = 0
horizontal tangents at { = 0, 2, 4.
i 0 ({) A 0 if { ? 0 or 2 ? { ? 4 i is increasing on (> 0) and (2> 4).
i 0 ({) ? 0 if 0 ? { ? 2 or { A 4 i is decreasing on (0> 2) and (4> ).
i 00 ({) A 0 if 1 ? { ? 3 i is concave upward on (1> 3).
i 00 ({) ? 0 if { ? 1 or { A 3 i is concave downward on (> 1)
and (3> ). There are inection points when { = 1 and 3.
22. i 0 (1) = i 0 (1) = 0
horizontal tangents at { = ยฑ1.
0
i ({) ? 0 if |{| ? 1 i is decreasing on (1> 1).
i 0 ({) A 0 if 1 ? |{| ? 2 i is increasing on (2> 1) and (1> 2).
i 0 ({) = 1 if |{| A 2 the graph of i has constant slope 1 on (> 2)
and (2> ).
i 00 ({) ? 0 if 2 ? { ? 0 i is concave downward on (2> 0). The point (0> 1) is an inection point.
23. i 0 ({) A 0 if |{| ? 2
i is increasing on (2> 2).
i 0 ({) ? 0 if |{| A 2 i is decreasing on (> 2)
and (2> ).
i 0 (2) = 0 horizontal tangent at
{ = 2. lim |i 0 ({)| = there is a vertical
{2
asymptote or vertical tangent (cusp) at { = 2. i 00 ({) A 0 if { 6= 2 i is concave upward on (> 2) and (2> ).
24. i 0 ({) A 0 if |{| ? 2
i is increasing on (2> 2). i 0 ({) ? 0 if |{| A 2
i is decreasing on (> 2) and (2> ). i 0 (2) = 0, so i has a horizontal tangent
(and local maximum) at { = 2. lim i ({) = 1 | = 1 is a horizontal asymptote.
{
i ({) = i ({) i is an odd function (its graph is symmetric about the origin).
Finally, i 00 ({) ? 0 if 0 ? { ? 3 and i 00 ({) A 0 if { A 3, so i is CD on (0> 3) and
CU on ((3> ).
)
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VE
2
2
25. (a) Since h{ is positive for all {, i 0 ({) = {h{ is positive where { A 0 and negative where { ? 0. Thus, i is increasing
on (0> ) and decreasing on (> 0).
(b) Since i changes from decreasing to increasing at { = 0, i has a minimum value there.
2
26. Since i 0 ({) = h{ A 0 on R, i is increasing on R.
27. (a) To nd the intervals on which i is increasing, we need to nd the intervals on which i 0 ({) = 3{2 1 is positive.
3{2 1 A 0 3{2 A 1 {2 A 13
|{| A
t
1
, so {
3
t
> 13
t
. Thus, i is
1
>
3
t
t
t t
1
>
.
In
a
similar
fashion,
i
is
decreasing
on
13 > 13 .
increasing on > 13 and on
3
(b) To nd the intervals on which i is concave upward, we need to nd the intervals on which i 00 ({) = 6{ is positive.
6{ A 0 { A 0. So i is concave upward on (0> ) and i is concave downward on (> 0).
(c) There is an inection point at (0> 0) since i changes its direction of concavity at { = 0.
i ({ + k) i ({)
[({ + k)4 2({ + k)2 ] ({4 2{2 )
= lim
k0
k0
k
k
4
3
2 2
3
4
2
({ + 4{ k + 6{ k + 4{k + k 2{ 4{k 2k2 ) ({4 2{2 )
= lim
k0
k
4{3 k + 6{2 k2 + 4{k3 + k4 4{k 2k2
= lim
= lim (4{3 + 6{2 k + 4{k2 + k3 4{ 2k) = 4{3 4{
k0
k0
k
28. (a) i 0({) = lim
i 0 ({ + k) i 0 ({)
[4({ + k)3 4({ + k)] (4{3 4{)
= lim
k0
k0
k
k
3
2
2
3
(4{ + 12{ k + 12{k + 4k 4{ 4k) (4{3 4{)
12{2 k + 12{k2 + 4k3 4k
= lim
= lim
k0
k0
k
k
= lim (12{2 + 12{k + 4k2 4) = 12{2 4
i 00({) = lim
k0
(b) i 0 ({) A 0 4{3 4{ A 0 4{ {2 1 A 0 4{({ + 1)({ 1) A 0, so i is increasing on (1> 0)
and (1> ) and i is decreasing on (> 1) and (0> 1).
(c) i 00 ({) A 0 12{2 4 A 0 12{2 A 4 {2 A 13
t
1
>
3
|{| A
t
1
, so i is CU on
3
t
> 13 and
t t
and i is CD on 13 > 13 .
29. e is the antiderivative of i . For small {, i is negative, so the graph of its antiderivative must be decreasing. But both d and f
are increasing for small {, so only e can be i โs antiderivative. Also, i is positive where e is increasing, which supports our
conclusion.
30. We know right away that f cannot be i โs antiderivative, since the slope of f is not zero at the {-value where i = 0. Now i is
positive when d is increasing and negative when d is decreasing, so d is the antiderivative of i .
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SECTION 22.88 WHAT D
DOES i 0 SAY ABOUT i ?
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The graph of I must start at (0> 1). Where the given graph, | = i ({), has a
31.
local minimum or maximum, the graph of I will have an inection point.
Where i is negative (positive), I is decreasing (increasing).
Where i changes from negative to positive, I will have a minimum.
Where i changes from positive to negative, I will have a maximum.
Where i is decreasing (increasing), I is concave downward (upward).
Where y is positive (negative), v is increasing (decreasing).
32.
Where y is increasing (decreasing), v is concave upward (downward).
Where y is horizontal (a steady velocity), v is linear.
33. i ({) =
sin {
, 2 { 2
1 + {2
Note that the graph of i is one of an odd function, so the graph of I will
be one of an even function.
34. i ({) =
{4 2{2 + 2 2, 3 { 3
Note that the graph of i is one of an even
function, so the graph of I will be one of an
odd function.
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CHAPTER 2 LIMITS AND
ND DERIVATIVES
VE
2 Review
1. (a) lim i ({) = O: See Denition 2.2.1 and Figures 1 and 2 in Section 2.2.
{d
(b) lim i ({) = O: See the paragraph after Denition 2.2.2 and Figure 9(b) in Section 2.2.
{d+
(c) lim i({) = O: See Denition 2.2.2 and Figure 9(a) in Section 2.2.
{d
(d) lim i ({) = : See Denition 2.5.1 and Figure 2 in Section 2.5.
{d
(e) lim i ({) = O: See Denition 2.5.4 and Figure 9 in Section 2.5.
{
2. In general, the limit of a function fails to exist when the function does not approach a xed number. For each of the following
functions, the limit fails to exist at { = 2.
The left- and right-hand
There is an
There are an innite
limits are not equal.
innite discontinuity.
number of oscillations.
3. (a) โ (g) See the statements of Limit Laws 1โ 6 and 11 in Section 2.3.
4. See Theorem 3 in Section 2.3.
5. (a) See Denition 2.5.2 and Figures 2 โ 4 in Section 2.5.
(b) See Denition 2.5.5 and Figures 9 and 10 in Section 2.5.
6. (a) | = {4 : No asymptote
(b) | = sin {: No asymptote
(c) | = tan {: Vertical asymptotes { = 2 + q, q an integer
(d) | = h{ : Horizontal asymptote | = 0
lim h{ = 0
(e) | = ln {: Vertical asymptote { = 0
{
lim ln { =
{0+
(f ) | = 1@{: Vertical asymptote { = 0, horizontal asymptote | = 0
(g) | = {: No asymptote
7. (a) A function i is continuous at a number d if i({) approaches i (d) as { approaches d; that is, lim i ({) = i (d).
{d
(b) A function i is continuous on the interval (> ) if i is continuous at every real number d. The graph of such a
function has no breaks and every vertical line crosses it.
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153
8. See Theorem 2.4.10.
9. See Denition 2.6.1.
10. See the paragraph containing Formula 3 in Section 2.6.
11. (a) The average rate of change of | with respect to { over the interval [{1 > {2 ] is
(b) The instantaneous rate of change of | with respect to { at { = {1 is lim
{2 {1
i ({2 ) i({1 )
.
{2 {1
i ({2 ) i ({1 )
.
{2 {1
12. See Denition 2.7.2. The pages following the denition discuss interpretations of i 0 (d) as the slope of a tangent line to the
graph of i at { = d and as an instantaneous rate of change of i ({) with respect to { when { = d.
13. See the paragraphs before and after Example 7 in Section 2.7.
14. (a) A function i is differentiable at a number d if its derivative i 0 exists
(c)
0
at { = d; that is, if i (d) exists.
(b) See Theorem 2.7.4. This theorem also tells us that if i is not
continuous at d, then i is not differentiable at d.
15. See the discussion and Figure 8 on page 152.
16. (a) See the rst box in Section 2.8.
(b) See the second box in Section 2.8.
17. (a) An antiderivative of a function i is a function I such that I 0 = i.
(b) The antiderivative of a velocity function is a position function (the derivative of a position function is a velocity function).
The antiderivative of an acceleration function is a velocity function (the derivative of a velocity function is an acceleration
function).
1. False.
Limit Law 2 applies only if the individual limits exist (these donโt).
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. True.
The limit doesnโt exist since i ({)@j({) doesnโt approach any real number as { approaches 5.
(The denominator approaches 0 and the numerator doesnโt.)
5. False.
6. False.
Consider lim
{5
{({ 5)
sin({ 5)
or lim
. The rst limit exists and is equal to 5. By Example 3 in Section 2.2,
{5
{5
{5
we know that the latter limit exists (and it is equal to 1).
1
. It exists (its value is 1) but i (6) = 0 and j(6) does not exist,
Consider lim [i({)j({)] = lim ({ 6)
{6
{6
{6
so i (6)j(6) 6= 1.
7. True.
A polynomial is continuous everywhere, so lim s({) exists and is equal to s(e).
{e
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CHAPTER 2 LIMITS AND
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VE
8. False.
Consider lim [i({) j({)] = lim
{0
{0
1
1
4 . This limit is (not 0), but each of the individual functions
{2
{
approaches .
9. True.
See Figure 11 in Section 2.5.
10. False.
Consider i ({) = sin { for { 0. lim i ({) 6= ยฑ and i has no horizontal asymptote.
{
+
1@({ 1) if { 6= 1
11. False.
Consider i ({) =
12. False.
The function i must be continuous in order to use the Intermediate Value Theorem. For example, let
+
1
if 0 { ? 3
There is no number f [0> 3] with i (f) = 0.
i ({) =
1 if { = 3
13. True.
Use Theorem 2.4.8 with d = 2, e = 5, and j({) = 4{2 11. Note that i (4) = 3 is not needed.
14. True.
Use the Intermediate Value Theorem with d = 1, e = 1, and Q = , since 3 ? ? 4.
15. False.
See the note after Theorem 4 in Section 2.7.
16. True.
i 0(u) exists i is differentiable at u
17. False.
18. False.
2
if { = 1
i is continuous at u
lim i ({) = i (u).
{u
2
g|
g 2|
is
the
second
derivative
while
is the rst derivative squared. For example, if | = {,
g{2
g{
2
g|
g 2|
=
0,
but
= 1.
then
g{2
g{
+2
{ + 1 if { 6= 0
For example, let i ({) =
2
if { = 0
Then i({) A 1 for all {, but lim i ({) = lim {2 + 1 = 1.
{0
{0
1. (a) (i) lim i ({) = 3
(ii)
{2+
lim i ({) = 0
{3+
(iii) lim i ({) does not exist since the left and right limits are not equal. (The left limit is 2.)
{3
(iv) lim i ({) = 2
{4
(v) lim i ({) =
(vi) lim i ({) =
(vii) lim i ({) = 4
(viii) lim i ({) = 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) i is discontinuous at { = 3, 0, 2, and 4. The discontinuities are jump, innite, innite, and removable, respectively.
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2.
lim i({) = 2,
{
lim i ({) = 0,
{
lim i ({) = ,
CHAPTER 2 REVIEW
ยค
lim i ({) = ,
{3
lim i ({) = 2,
{3
{3+
i is continuous from the right at 3
3
3. Since the exponential function is continuous, lim h{ { = h11 = h0 = 1.
{1
{2 9
32 9
0
=
=
= 0.
{3 {2 + 2{ 3
32 + 2(3) 3
12
4. Since rational functions are continuous, lim
{2 9
({ + 3)({ 3)
{3
3 3
6
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 + 2{ 3
{1+ { + 2{ 3
3
k 3k2 + 3k 1 + 1
(k 1)3 + 1
k3 3k2 + 3k
7. lim
= lim
= lim
= lim k2 3k + 3 = 3
k0
k0
k0
k0
k
k
k
6. lim
Another solution: Factor the numerator as a sum of two cubes and then simplify.
[(k 1) + 1] (k 1)2 1(k 1) + 12
(k 1)3 + 1
(k 1)3 + 13
= lim
= lim
lim
k0
k0
k0
k
k
k
2
= lim (k 1) k + 2 = 1 0 + 2 = 3
k0
w2 4
(w + 2)(w 2)
w+2
2+2
4
1
= lim
= lim 2
=
=
=
w2 w3 8
w2 (w 2)(w2 + 2w + 4)
w2 w + 2w + 4
4+4+4
12
3
8. lim
u
9. lim
= since (u 9)4
u9 (u 9)4
10. lim
4y
y4+ |4 y|
= lim
4y
y4+ (4 y)
0 as u
= lim
1
y4+ 1
u
9 and
A 0 for u 6= 9.
(u 9)4
= 1
x4 1
(x2 + 1)(x2 1)
(x2 + 1)(x + 1)(x 1)
(x2 + 1)(x + 1)
2(2)
4
=
lim
=
lim
=
lim
=
=
x1 x3 + 5×2 6x
x1 x(x2 + 5x 6)
x1
x1
x(x + 6)(x 1)
x(x + 6)
1(7)
7
11. lim
{+6{
{+6{
{+6+{
( { + 6 )2 {2
ยท
=
lim
=
lim
{3 {3 3{2
{3
{3 {2 ({ 3)
{2 ({ 3)
{+6+{
{+6+{
12. lim
({2 { 6)
({ 3)({ + 2)
{ + 6 {2
= lim
= lim
{3 {2 ({ 3)
{3 {2 ({ 3)
{3 {2 ({ 3)
{+6+{
{+6+{
{+6+{
= lim
= lim
{3 {2
13. Let w = sin {. Then as {
14.
({ + 2)
5
5
=
=
9(3
+
3)
54
{+6+{
, sin {
0+ , so w
0+ . Thus, lim ln(sin {) = lim ln w = .
{
w0+
1 2{2 {4
(1 2{2 {4 )@{4
1@{4 2@{2 1
001
1
1
= lim
= lim
=
=
=
4
4
4
{ 5 + { 3{
{ (5 + { 3{ )@{
{ 5@{4 + 1@{3 3
0+03
3
3
lim
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CHAPTER 2 LIMITS AND
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VE
15. Since { is positive,
{2 = |{| = {. Thus,
s
1 9@{2
{2 9
{2 9@ {2
10
1
lim
= lim
= lim
=
=
{ 2{ 6
{ (2{ 6)@{
{
2 6@{
20
2
16. Let w = { {2 = {(1 {). Then as {
, w
2
, and lim h{{ = lim hw = 0.
{
w
2
{ + 4{ + 1 {
{2 + 4{ + 1 + {
({2 + 4{ + 1) {2
2
ยท
17. lim
{ + 4{ + 1 { = lim
= lim
{
{
{
1
{2 + 4{ + 1 + {
{2 + 4{ + 1 + {
l
k
(4{ + 1)@{
= lim
divide by { = {2 for { A 0
{ (
{2 + 4{ + 1 + {)@{
4 + 1@{
4
4+0
= lim s
=
= =2
{
2
1+0+0+1
1 + 4@{ + 1@{2 + 1
18. lim
{1
1
1
+ 2
{1
{ 3{ + 2
1
{2
1
1
+
+
= lim
= lim
{1 { 1
{1 ({ 1)({ 2)
({ 1)({ 2)
({ 1)({ 2)
1
{1
1
= lim
= lim
=
= 1
{1 ({ 1)({ 2)
{1 { 2
12
19. 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
0
cos2 {
1
2
{2
{2
{
0
cos2 {
1
2 . But lim 0 = 0 and
{ยฑ
{2
{
1
cos2 {
=
0,
so
by
the
Squeeze
Theorem,
lim
= 0.
{ยฑ {2
{ยฑ
{2
lim
cos2 {
= because cos2 {
{0
{2
Thus, | = 0 is the horizontal asymptote. lim
1 and {2
0 as {
0, so { = 0 is the
vertical asymptote.
20. From the graph of | = i ({) =
{2 + { + 1 {2 {, it appears that there are 2 horizontal asymptotes and possibly 2
vertical asymptotes. To obtain a different form for i , letโs multiply and divide it by its conjugate.
{2 + { + 1 + {2 {
({2 + { + 1) ({2 {)
2
2
=
i1 ({) =
{ +{+1 { {
{2 + { + 1 + {2 {
{2 + { + 1 + {2 {
2{ + 1
=
2
{ + { + 1 + {2 {
Now
2{ + 1
lim i1 ({) = lim
{
{2 + { + 1 + {2 {
{
2 + (1@{)
s
= lim s
{
1 + (1@{) + (1@{2 ) + 1 (1@{)
=
[since
{2 = { for { A 0]
2
= 1,
1+1
so | = 1 is a horizontal asymptote. For { ? 0, we have
{2 = |{| = {, so when we divide the denominator by {,
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157
with { ? 0, we get
&
%u
u
{2 + { + 1 + {2 {
{2 + { + 1 + {2 {
1
1
1
=
=
1+ + 2 + 1
{
{
{
{
{2
Therefore,
2{ + 1
2 + (1@{)
ks
l
= lim
lim i1 ({) = lim
s
2
2
{
{
{ + { + 1 + { { {
1 + (1@{) + (1@{2 ) + 1 (1@{)
=
2
= 1>
(1 + 1)
so | = 1 is a horizontal asymptote.
The domain of i is (> 0]
1, so
0 , i ({)
3, so { = 1
1+ , i ({)
[1> ). As {
{ = 0 is not a vertical asymptote. As {
is not a vertical asymptote and hence there are no vertical asymptotes.
21. Since 2{ 1 i({) {2 for 0 ? { ? 3 and lim (2{ 1) = 1 = lim {2 , we have lim i ({) = 1 by the Squeeze Theorem.
{1
{1
{1
22. Let i({) = {2 , j({) = {2 cos 1@{2 and k({) = {2 . Then since cos 1@{2 1 for { 6= 0, we have
i ({) j({) k({) for { 6= 0, and so lim i ({) = lim k({) = 0
{0
23. (a) i ({) =
{0
lim j({) = 0 by the Squeeze Theorem.
{0
{ if { ? 0, i ({) = 3 { if 0 { ? 3, i ({) = ({ 3)2 if { A 3.
{ = 0
(i) lim i ({) = lim (3 {) = 3
(ii) lim i ({) = lim
(iii) Because of (i) and (ii), lim i ({) does not exist.
(iv) lim i ({) = lim (3 {) = 0
(v) lim i ({) = lim ({ 3)2 = 0
(vi) Because of (iv) and (v), lim i ({) = 0.
{0+
{0+
{0
{0
{3+
{3
{0
{3
{3
{3+
(b) i is discontinuous at 0 since lim i ({) does not exist.
(c)
{0
i is discontinuous at 3 since i (3) does not exist.
24. (a) {2 9 is continuous on R since it is a polynomial and
continuous on { | {2 9 0 = (> 3]
{ is continuous on [0> ), so the composition {2 9 is
[3> ). Note that {2 2 6= 0 on this set and so the quotient function
{2 9
j({) = 2
is continuous on its domain, (> 3]
{ 2
[3> ).
(b) sin { is continuous on R by Theorem 7 in Section 2.5. Since h{ is continuous on R, hsin { is continuous on R by
Theorem 9 in Section 2.5. Lastly, { is continuous on R since itโs a polynomial and the product {hsin { is continuous on its
domain R by Theorem 4 in Section 2.5.
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VE
25. i ({) = 2{3 + {2 + 2 is a polynomial, so it is continuous on [2> 1] and i (2) = 10 ? 0 ? 1 = i(1). So by the
Intermediate Value Theorem there is a number f in (2> 1) such that i (f) = 0, that is, the equation 2{3 + {2 + 2 = 0 has a
root in (2> 1).
2
26. i ({) = h{ { is continuous on R so it is continuous on [0> 1]. i(0) = 1 A 0 A 1@h 1 = i (1). So by the Intermediate
2
2
Value Theorem, there is a number f in (0> 1) such that i(f) = 0. Thus, h{ { = 0> or h{ = {, has a root in (0> 1).
27. (a) v = v(w) = 1 + 2w + w2 @4. The average velocity over the time interval [1> 1 + k] is
1 + 2(1 + k) + (1 + k)2 4 13@4
10k + k2
10 + k
v(1 + k) v(1)
=
=
=
yave =
(1 + k) 1
k
4k
4
So for the following intervals the average velocities are:
(i) [1> 3]: k = 2, yave = (10 + 2)@4 = 3 m@s
(ii) [1> 2]: k = 1, yave = (10 + 1)@4 = 2=75 m@s
(iii) [1> 1=5]: k = 0=5, yave = (10 + 0=5)@4 = 2=625 m@s
(iv) [1> 1=1]: k = 0=1, yave = (10 + 0=1)@4 = 2=525 m@s
(b) When w = 1, the instantaneous velocity is lim
k0
v(1 + k) v(1)
10 + k
10
= lim
=
= 2=5 m@s.
k0
k
4
4
28. (a) When Y increases from 200 in3 to 250 in3 , we have Y = 250 200 = 50 in3 , and since S = 800@Y ,
S = S (250) S (200) =
is
800
800
= 3=2 4 = 0=8 lb@in2 . So the average rate of change
250
200
S
0=8
lb@in2
.
=
= 0=016
Y
50
in3
(b) Since Y = 800@S , the instantaneous rate of change of Y with respect to S is
lim
Y
k0 S
= lim
k0
= lim
Y (S + k) Y (S )
800@(S + k) 800@S
800 [S (S + k)]
= lim
= lim
k0
k0
k
k
k(S + k)S
800
k0 (S + k)S
=
800
S2
which is inversely proportional to the square of S .
29. Estimating the slopes of the tangent lines at { = 2, 3, and 5, we obtain approximate values 0=4, 2, and 0=1. Since the
graph is concave downward at { = 5, i 00 (5) is negative. Arranging the numbers in increasing order, we have:
i 00 (5) ? 0 ? i 0 (5) ? i 0 (2) ? 1 ? i 0 (3).
i({) i (2)
{3 2{ 4
= lim
{2
{2
{2
{2
30. (a) i 0 (2) = lim
(c)
({ 2)({2 + 2{ + 2)
= lim ({2 + 2{ + 2) = 10
{2
{2
{2
= lim
(b) | 4 = 10({ 2) or | = 10{ 16
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31. (a) Estimating i 0 (1) from the triangle in the graph,
|
0=37
= 0=74.
{
0=50
To estimate i 0 (1) numerically, we have
we get
2
i (1 + k) i (1)
h(1+k) h1
i (1) = lim
= lim
=|
k0
k0
k
k
0
From the table, we have i 0 (1)
(b) | h1
0=736({ 1) or |
0=736.
0=736{ + 1=104
(c) See the graph in part (a).
k
|
0=01
0=001
0=0001
0=01
0=001
0=0001
0=732
0=735
0=736
0=739
0=736
0=736
32. 26 = 64, so i ({) = {6 and d = 2.
33. (a) i 0 (u) 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 u increases, F increases. So i 0 (u) will always be positive.
34.
35.
i({ + k) i ({)
= lim
37. (a) i ({) = lim
k0
k0
k
0
= lim
k0
36.
s
s
3 5({ + k) 3 5{ 3 5({ + k) + 3 5{
s
k
3 5({ + k) + 3 5{
[3 5({ + k)] (3 5{)
5
5
s
= lim s
=
k0
2
3
5{
3 5({ + k) + 3 5{
k
3 5({ + k) + 3 5{
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CHAPTER 2 LIMITS AND
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VE
(b) Domain of i : (the radicand must be nonnegative) 3 5{ 0
5{ 3 {
> 35
Domain of i 0 : exclude 35 because it makes the denominator zero;
{
> 35
(c) Our answer to part (a) is reasonable because i 0 ({) is always negative and i
is always decreasing.
38. (a) As {
ยฑ, i ({) = (4 {)@(3 + {)
asymptote at | = 1. As {
i ({)
1, so there is a horizontal
3+ , i ({)
, and as {
3 ,
. Thus, there is a vertical asymptote at { = 3.
(b) Note that i is decreasing on (> 3) and (3> ), so i 0 is negative on
those intervals. As {
i0
ยฑ, i 0
0. As {
3 and as {
3+ ,
.
4 ({ + k)
4{
i ({ + k) i ({)
3 + ({ + k)
3+{
(3 + {) [4 ({ + k)] (4 {) [3 + ({ + k)]
(c) i 0 ({) = lim
= lim
= lim
k0
k0
k0
k
k
k [3 + ({ + k)] (3 + {)
(12 3{ 3k + 4{ {2 k{) (12 + 4{ + 4k 3{ {2 k{)
k0
k[3 + ({ + k)](3 + {)
= lim
= lim
7k
k0 k [3 + ({ + k)] (3 + {)
= lim
7
k0 [3 + ({ + k)] (3 + {)
=
7
(3 + {)2
(d) The graphing device conrms our graph in part (b).
39. i is not differentiable: at { = 4 because i is not continuous, at { = 1 because i has a corner, at { = 2 because i is not
continuous, and at { = 5 because i has a vertical tangent.
40. The graph of d has tangent lines with positive slope for { ? 0 and negative slope for { A 0, and the values of f t this pattern,
so f must be the graph of the derivative of the function for d. The graph of f has horizontal tangent lines to the left and right of
the {-axis and e has zeros at these points. Hence, e is the graph of the derivative of the function for f. Therefore, d is the graph
of i , f is the graph of i 0 , and e is the graph of i 00 .
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161
41. F 0 (1990) is the rate at which the total value of US currency in circulation is changing in billions of dollars per year. To
estimate the value of F 0 (1990), we will average the difference quotients obtained using the times w = 1985 and w = 1995.
Let D =
E=
187=3 271=9
84=6
F(1985) F(1990)
=
=
= 16=92 and
1985 1990
5
5
409=3 271=9
137=4
F(1995) F(1990)
=
=
= 27=48. Then
1995 1990
5
5
F 0 (1990) = lim
w1990
F(w) F(1990)
w 1990
D+E
16=92 + 27=48
44=4
=
=
= 22=2 billion dollars@year.
2
2
2
42. Let F(w) be the function that denotes the cost of living in terms of time w. F(w) is an increasing function, so F 0 (w) A 0. Since
the cost of living is rising at a slower rate, the slopes of the tangent lines are positive but decreasing as w increases. Hence,
F 00 (w) ? 0.
43. (a) i 0 ({) A 0 on (2> 0) and (2> )
i is increasing on those intervals. i 0 ({) ? 0 on (> 2) and (0> 2)
i is decreasing on those intervals.
(b) i 0 ({) = 0 at { = 2, 0, and 2, so these are where local maxima or minima will occur. At { = ยฑ2, i 0 changes from
negative to positive, so i has local minima at those values. At { = 0, i 0 changes from positive to negative, so i has a local
maximum there.
(c) i 0 is increasing on (> 1) and (1> )
(d)
i 00 A 0 and i is concave upward on those intervals.
i 0 is decreasing on (1> 1) i 00 ? 0 and
i is concave downward on this interval.
44. (a)
(b)
45. i (0) = 0, i 0 (2) = i 0 (1) = i 0 (9) = 0, lim i ({) = 0, lim i({) = ,
{
0
{6
0
i ({) ? 0 on (> 2), (1> 6), and (9> ), i ({) A 0 on (2> 1) and (6> 9),
i 00 ({) A 0 on (> 0) and (12> ), i 00 ({) ? 0 on (0> 6) and (6> 12)
46. (a) Drawing slope triangles, we obtain the following estimates: I 0 (1950)
and I 0 (1987)
1=1
= 0=11, I 0 (1965)
10
1=6
= 0=16,
10
0=2
= 0=02.
10
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CHAPTER 2 LIMITS AND
ND DERIVATIVES
VE
(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:
โข In the baby-boom era (post-WWII), there was optimism about the economy and family size was rising.
โข In the baby-bust era, there was less economic optimism, and it was considered less socially responsible to have a
large family.
โข In the baby-boomlet era, there was increased economic optimism and a return to more conservative attitudes.
47. (a) Using the data closest to w = 6, we have
and
v(8) v(6)
180 95
=
= 42=5
86
2
v(4) v(6)
40 95
=
= 27=5. Averaging these two values gives us
46
2
42=5 + 27=5
= 35 ft@s as an estimate for the speed of the car after
2
6 seconds.
(b) From the graph, it appears that the inection point is at (8> 180).
(c) The velocity of the car is at a maximum at the inection point.
48. Let i be the function shown. Since i is negative for { ? 0 and positive for { A 0,
I is decreasing for { ? 0 and increasing for { A 0. i is increasing on (d> d)
(from the low point to the high point) so its derivative i 0 (the second derivative of I )
is positive, making I concave upward on (d> d). i is decreasing elsewhere, so its
derivative i 0 is negative and I is concave downward on (> d) and (d> ).
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FOCUS ON PROBLEM SOLVING
1. Let w = 6 {, so { = w6 . Then w
1 as {
1, so
3
{1
w2 1
(w 1)(w + 1)
w+1
1+1
2
= lim
= lim 2
= 2
= .
lim
= lim 3
2
{1
w1
w1
w1
w 1
(w 1) (w + w + 1)
w +w+1
1 +1+1
3
{1
3
3
Another method: Multiply both the numerator and the denominator by ( { + 1)
{2 + { + 1 .
d{ + e 2
d{ + e + 2
d{ + e 4
. Now since the denominator
ยท
= lim
{0
{
d{ + e + 2 {0 { d{ + e + 2
2. First rationalize the numerator: lim
approaches 0 as {
0, the limit will exist only if the numerator also approaches 0 as {
0. So we require that
d
d
=1
d(0) + e 4 = 0 e = 4. So the equation becomes lim
= 1 d = 4.
{0
d{ + 4 + 2
4+2
Therefore, d = e = 4.
3. For 12 ? { ? 12 , we have 2{ 1 ? 0 and 2{ + 1 A 0, so |2{ 1| = (2{ 1) and |2{ + 1| = 2{ + 1.
|2{ 1| |2{ + 1|
(2{ 1) (2{ + 1)
4{
= lim
= lim
= lim (4) = 4.
{0
{0
{0
{
{
{
1 1 2
4. Let U be the midpoint of RS , so the coordinates of U are 2 {> 2 { since the coordinates of S are {> {2 . Let T = (0> d).
Therefore, lim
{0
2
1
{ d
{2 2d
{2
1
=
= {, pTU = (negative reciprocal). But pTU = 21
, so we conclude that
{
{
{
{0
2
1
1 = {2 2d 2d = {2 + 1 d = 12 {2 + 12 . As {
0, d
> and the limiting position of T is 0> 12 .
2
Since the slope pRS =
{
[[{]] + 1
{
1
[[{]]
?
1
?1+
for { 1. As {
[[{]]
[[{]]
[[{]]
[[{]]
[[{]]
{
1. Thus, lim
= 1 by the Squeeze Theorem.
{ [[{]]
5. Since [[{]] { ? [[{]] + 1, we have
so
1
[[{]]
0 and 1 +
2
1
[[{]]
2
, [[{]]
,
2
6. (a) [[{]] + [[|]] = 1. Since [[{]] and [[|]]2 are positive integers or 0, there are
only 4 cases:
Case (i): [[{]] = 1, [[|]] = 0
1 { ? 2 and 0 | ? 1
Case (ii): [[{]] = 1, [[|]] = 0 1 { ? 0 and 0 | ? 1
Case (iii):[[{]] = 0, [[|]] = 1
0 { ? 1 and 1 | ? 2
Case (iv): [[{]] = 0, [[|]] = 1 0 { ? 1 and 1 | ? 0
(b) [[{]]2 [[|]]2 = 3. The only integral solution of q2 p2 = 3 is q = ยฑ2
and p = ยฑ1. So the graph is
,
2 { 3 or 2 { ? 1>
.
({> |)
1 | ? 2 or 1 | ? 0
+
{({> |) | [[{]] = ยฑ2, [[|]] = ยฑ1} =
(c) [[{ + |]]2 = 1 [[{ + |]] = ยฑ1 1 { + | ? 2
or 1 { + | ? 0
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CHAPTER 2 FOCUS ON PROBLEM
MS
SOLVING
(d) For q { ? q + 1, [[{]] = q. Then [[{]] + [[|]] = 1 [[|]] = 1 q
1 q | ? 2 q. Choosing integer values for q produces the graph.
7. i is continuous on (> d) and (d> ). To make i continuous on R, we must have continuity at d. Thus,
lim {2 = lim ({ + 1) d2 = d + 1 d2 d 1 = 0
{d
[by the quadratic formula] d = 1 ยฑ 5 2 1=618 or 0=618.
lim i ({) = lim i ({)
{d+
{d
{d+
8. (a) Here are a few possibilities:
(b) The โobstacleโ is the line { = | (see diagram). Any intersection of the graph of i with the line | = { constitutes a xed
point, and if the graph of the function does not cross the line somewhere in (0> 1), then it must either start at (0> 0)
(in which case 0 is a xed point) or nish at (1> 1) (in which case 1 is a xed point).
(c) Consider the function I ({) = i({) {, where i is any continuous function with domain [0> 1] and range in [0> 1]. We
shall prove that i has a xed point. Now if i (0) = 0 then we are done: i has a xed point (the number 0), which is what
we are trying to prove. So assume i (0) 6= 0. For the same reason we can assume that i (1) 6= 1. Then I (0) = i (0) A 0
and I (1) = i(1) 1 ? 0. So by the Intermediate Value Theorem, there exists some number f in the interval (0> 1) such
that I (f) = i (f) f = 0. So i (f) = f, and therefore i has a xed point.
9. (a) Consider J({) = W ({ + 180 ) W ({). Fix any number d. If J(d) = 0, we are done: Temperature at d = Temperature
at d + 180 . If J(d) A 0, then J(d + 180 ) = W (d + 360 ) W (d + 180 ) = W (d) W (d + 180 ) = J(d) ? 0.
Also, J is continuous since temperature varies continuously. So, by the Intermediate Value Theorem, J has a zero on the
interval [d> d + 180 ]. If J(d) ? 0, then a similar argument applies.
(b) Yes. The same argument applies.
(c) The same argument applies for quantities that vary continuously, such as barometric pressure. But one could argue that
altitude above sea level is sometimes discontinuous, so the result might not always hold for that quantity.
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165
10. (a) Solution 1: We introduce a coordinate system and drop a perpendicular
from S , as shown. We see from QFS that tan 2 =
|
, and from
1{
QES that tan = |@{ . Using the double-angle formula for tangents,
we get
|
2 tan
2( |@{)
= tan 2 =
=
. After a bit of
2
1{
1 tan
1 ( |@{)2
simplication, this becomes
2{
1
= 2
1{
{ |2
| 2 = { (3{ 2).
As the altitude DP decreases in length, the point S will approach the {-axis, that is, |
0, so the limiting location of S
must be one of the roots of the equation {(3{ 2) = 0. Obviously it is not { = 0 (the point S can never be to the left of
the altitude DP , which it would have to be in order to approach 0) so it must be 3{ 2 = 0, that is, { = 23 .
Solution 2: We add a few lines to the original diagram, as shown. Now note
that ES T = S EF (alternate angles; TS k EF by symmetry) and
similarly FTS = TFE. So ES T and FTS are isosceles, and
the line segments ET, TS and S F are all of equal length. As |DP |
0,
S and T approach points on the base, and the point S is seen to approach a
position two-thirds of the way between E and F, as above.
(b) The equation | 2 = {(3{ 2) calculated in part (a) is the equation of
the curve traced out by S . Now as |DP|
{
1, and since tan = |@{, |
, 2
,
2
,
4
1. Thus, S only traces out the
part of the curve with 0 | ? 1.
11. Let d be the {-coordinate of T. Since the derivative of | = 1 {2 is | 0 = 2{, the slope at T is 2d. But since the triangle
3@1, so the slope at T is 3. Therefore, we must have that 2d = 3 d = 23 .
2
Thus, the point T has coordinates 23 > 1 23
= 23 > 14 and by symmetry, S has coordinates 23 > 14 .
is equilateral, DR@RF =
12. (a) Y 0 (w) is the rate of change of the volume of the water with respect to time. K 0 (w) is the rate of change of the height of the
water with respect to time. Since the volume and the height are increasing, Y 0 (w) and K 0 (w) are positive.
(b) Y 0 (w) is constant, so Y 00 (w) is zero (the slope of a constant function is 0).
(c) At rst, the height K of the water increases quickly because the tank is narrow. But as the sphere widens, the rate of
increase of the height slows down, reaching a minimum at w = w2 . Thus, the height is increasing at a decreasing rate on
(0> w2 ), so its graph is concave downward and K 00 (w1 ) ? 0. As the sphere narrows for w A w2 , the rate of increase of the
height begins to increase, and the graph of K is concave upward. Therefore, K 00 (w2 ) = 0 and K 00 (w3 ) A 0.
13. (a) Put { = 0 and | = 0 in the equation: i (0 + 0) = i(0) + i (0) + 02 ยท 0 + 0 ยท 02
Subtracting i(0) from each side of this equation gives i (0) = 0.
i (0) = 2i (0).
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CHAPTER 2 FOCUS ON PROBLEM
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SOLVING
i(0) + i (k) + 02 k + 0k2 i (0)
i (0 + k) i (0)
i (k)
i({)
= lim
= lim
= lim
=1
(b) i (0) = lim
{0
k0
k0
k0
k
k
k
{
i ({) + i (k) + {2 k + {k2 i ({)
i ({ + k) i ({)
i (k) + {2 k + {k2
0
= lim
= lim
(c) i ({) = lim
k0
k0
k0
k
k
k
i (k)
+ {2 + {k = 1 + {2
= lim
k0
k
0
14. We nd the equation of the parabola by substituting the point (100> 100), at which the car is situated, into the general
equation | = d{2 : 100 = d(100)2
1
d = 100
. Now we nd the equation of a tangent to the parabola at the point
1
1
1
(2{) = 50
{, so an equation of the tangent is | |0 = 50
{0 ({ {0 ).
({0 > |0 ). We can show that | 0 = d(2{) = 100
1
{20 , so our equation of the tangent can be simplied to
Since the point ({0 > |0 ) is on the parabola, we must have |0 = 100
1
1
| = 100
{20 + 50
{0 ({ {0 ). We want the statue to be located on the tangent line, so we substitute its coordinates (100> 50)
1
1
{20 + 50
{0 (100 {0 ) {20 200{0 + 5000 = 0
into this equation: 50 = 100
k
l
s
{0 = 12 200 ยฑ 2002 4(5000)
{0 = 100 ยฑ 50 2. But {0 ? 100, so the carโs headlights illuminate the statue
when it is located at the point 100 50 2> 150 100 2
(29=3> 8=6), that is, about 29=3 m east and 8=6 m north of
the origin.
15. lim i ({) = lim
{d
{d
1
1
2 [i ({) + j({)] + 2 [i({) j({)]
= 12 lim [i ({) + j({)] + 12 lim [i ({) j({)]
{d
{d
= 12 ยท 2 + 12 ยท 1 = 32 ,
and lim j({) = lim
{d
{d
[i ({) + j({)] i ({) = lim [i ({) + j({)] lim i({) = 2 32 = 12 .
{d
k
{d
lk
l
So lim [i ({)j({)] = lim i({) lim j({) = 32 ยท 12 = 34 .
{d
{d
{d
Another solution: Since lim [i ({) + j({)] and lim [i ({) j({)] exist, we must have
{d
lim [i ({) + j({)]2 =
{d
{d
2
2
lim [i ({) + j({)] and lim [i ({) j({)]2 = lim [i ({) j({)] , so
{d
{d
{d
[i ({) + j({)]2 [i ({) j({)]2
[because all of the i 2 and j2 cancel]
= 14 lim [i ({) + j({)]2 lim [i ({) j({)]2 = 14 22 12 = 34 .
lim [i ({) j({)] = lim 14
{d
{d
{d
{d
{i ({ + k) {i ({)
j({ + k) j({)
({ + k)i ({ + k) {i ({)
ki({ + k)
= lim
= lim
+
16. j ({) = lim
k0
k0
k0
k
k
k
k
0
= { lim
k0
i ({ + k) i ({)
+ lim i ({ + k) = {i 0 ({) + i ({)
k0
k
because i is differentiable and therefore continuous.
17. We are given that |i ({)| {2 for all {. In particular, |i (0)| 0, but |d| 0 for all d. The only conclusion is
2
{
i ({) i (0) i ({)
|i ({)|
{2
i ({) i (0)
=
= |{| |{|
|{|.
=
=
that i (0) = 0. Now
{0
{
|{|
|{|
|{|
{0
But lim (|{|) = 0 = lim |{|, so by the Squeeze Theorem, lim
{0
{0
{0
i ({) i (0)
= 0. So by the denition of a derivative,
{0
i is differentiable at 0 and, furthermore, i 0 (0) = 0.
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