Solution manual calculus 8th edition varberg, purcell, rigdon ch08

Indeterminate Forms and
Improper Integrals

8

CHAPTER

8.1 Concepts Review

7. The limit is not of the form

1. lim f ( x); lim g ( x)
x→a

2.

As x → 1– , x 2 – 2 x + 2 → 1, and x 2 – 1 → 0 – so

x →a

f ′( x)
g ′( x)

9. The limit is of the form

0
1. The limit is of the form .
0
2 x – sin x
2 – cos x
= lim

=1
lim
x
1
x →0
x →0

1

0
.
0
cos x
– sin x
lim
= lim
=1
x →π / 2 π / 2 – x x →π / 2 –1
0
.
0
1 – 2 cos 2 x

10. The limit is of the form

sin –1 x

=

=

1– 2
= –1
1

x2 + 6 x + 8

x → –2 x 2

– 3x –10
2
2
=
=–
–7
7

=

3
=3
1

x →0

476

x3 – 3 x 2 + x
x3 – 2 x

Section 8.1

11. The limit is of the form

lim

x →0

2x + 6
x → –2 2 x – 3

=

= lim

+

7

x

2

x

–1
–1

x →0

3x2 + 6 x + 1
3x2 – 2

0
.
0

7 x ln 7
2 x

= lim

x →0+ 2 x ln 2

= lim
x →0

2 x

+

7

x

ln 7

2

x

ln 2

ln 7
≈ 2.81
ln 2

13. The limit is of the form

= lim

0
.
0

1 – 2t
–3
t – t2
3
2 t
= lim
= 2 =–
lim
1
1
2
t →1 ln t
t →1

12. The limit is of the form

0
. (Apply l’Hôpital’s
0

Rule twice.)

0
6. The limit is of the form .
0

lim

0
.
0

t

0
5. The limit is of the form .
0

lim

–1

ex – e– x
ex + e– x 2
= lim

= =1
2
x →0 2sin x
x →0 2 cos x

0
.
0

3
1+ 9 x 2
lim
x →0 1
1– x 2

3sin 2 x cos x

lim

3. The limit is of the form

sec2 x

0
.
0

3
ln(sin x)3
= lim sin x

lim
x →π / 2 π / 2 – x
x →π / 2
0
=
=0
–1

2. The limit is of the form

tan –1 3 x

ln x 2

lim

Problem Set 8.1

4. The limit is of the form

0
.
0

1 2x
2
1
= lim x
= lim
=1

x →1 x 2 – 1 x →1 2 x
x →1 x 2

4. Cauchy’s Mean Value

x – sin 2 x
= lim
x →0 tan x
x →0

= –∞

8. The limit is of the form

x →0

x →0

x2 + 1

x →1–

3. sec2 x; 1; lim cos x ≠ 0

lim

x2 – 2 x + 2

lim

lim

0
.
0

lim

=

1
1
=–
–2
2

ln cos 2 x

x →0

= lim

7x

2

= lim

–2sin 2 x
cos 2 x

x →0

14 x

–4 cos 2 x

x →0 14 cos 2 x – 28 x sin 2 x

= lim

–2sin 2 x

x →0 14 x cos 2 x

=

–4
2
=–
14 – 0
7

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0

.
0
3sin x
3cos x
lim
= lim
1
–x
x →0 –
x →0 – –

14. The limit is of the form

19. The limit is of the form

Rule twice.)

2 –x

lim

x →0 –

0
. (Apply l’Hôpital’s
0

Rule three times.)
tan x – x
sec2 x – 1

lim
= lim
x →0 sin 2 x – 2 x x →0 2 cos 2 x – 2
2sec 2 x tan x
2sec 4 x + 4sec2 x tan 2 x
= lim
–8cos 2 x
x →0 –4sin 2 x
x →0
2+0
1
=
=–
–8
4
= lim

0
16. The limit is of the form . (Apply l’Hôpital’s
0
Rule three times.)
sin x – tan x
cos x – sec2 x
lim
= lim
x →0 x 2 sin x
x →0 2 x sin x + x 2 cos x
– sin x – 2sec2 x tan x
= lim
x →0 2sin x + 4 x cos x – x 2 sin x

– cos x – 2sec4 x – 4sec2 x tan 2 x
= lim
x →0
6 cos x – x 2 cos x – 6 x sin x
–1 – 2 – 0
1
=
=–
6–0–0
2
17. The limit is of the form

0
. (Apply l’Hôpital’s
0

Rule twice.)
x2
2x
2
lim
= lim
= lim
+ sin x – x
+ cos x – 1
+ − sin x
x →0
x →0
x →0
0

This limit is not of the form . As
0
x → 0+ , 2 → 2, and − sin x → 0− , so
2
lim
= −∞.
+ sin x
x →0

18. The limit is of the form

0
. (Apply l’Hôpital’s
0

Rule twice.)
e – ln(1 + x) –1

x →0

x
e +
x

= lim

x →0

8 x3

2

1
(1+ x )2

2

= lim

x →0

=

e

x

– 1+1x

= lim

1
1+ x 2

–1

24 x 2
1
1
= lim –

=–
24
x →0 24(1 + x 2 ) 2
x →0

20. The limit is of the form

= lim

–2 x
(1+ x 2 ) 2

x →0

48 x

0
. (Apply l’Hôpital’s
0

Rule twice.)
cosh x –1
sinh x
cosh x 1
= lim
= lim
=
lim
2`
2

x
2
2
0
0
x →0
x
→
x
→
x
21. The limit is of the form

0
. (Apply l’Hôpital’s
0

Rule twice.)
1 − cos x − x sin x
lim
2
+
x → 0 2 − 2 cos x − sin x
− x cos x
= lim
x → 0+ 2sin x − 2 cos x sin s
x sin x – cos x
= lim
2
2

+
x →0 2 cos x – 2 cos x + 2sin x
0
This limit is not of the form .
0
As x → 0+ , x sin x – cos x → −1 and
2 cos x – 2 cos 2 x + 2sin 2 x → 0+ , so
x sin x – cos x
lim
= –∞
+ 2 cos x – 2 cos 2 x + 2sin 2 x
x →0

22. The limit is of the form

lim

sin x + tan x
ex + e– x – 2

0
.
0

cos x + sec2 x

= lim

ex – e– x
0

This limit is not of the form .
0
x →0 –

x →0 –

As x → 0 – , cos x + sec 2 x → 2, and
e x – e – x → 0 – , so lim

cos x + sec2 x

x →0 –

23. The limit is of the form
x

x

lim

tan –1 x – x

x →0

= lim – 6 – x cos x = 0

15. The limit is of the form

0
. (Apply l’Hôpital’s

0

∫
lim 0

x →0

1 + sin t dt
x

ex – e– x

= – ∞.

0
.
0

= lim 1 + sin x = 1
x →0

2x

1+1
=1
2

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24. The limit is of the form
x

lim

x →0

∫0

t cos t dt
2

+

= lim
x →0

x
cos x

+

2 x

26. Note that sin (1 0 ) is undefined (not zero), so

0
.
0

= lim
x →0

+

l'Hôpital's Rule cannot be used.
1
⎛1⎞
As x → 0, → ∞ and sin ⎜ ⎟ oscillates rapidly
x
⎝ x⎠
between –1 and 1, so

x cos x
2x

=∞

lim

x →0

25. It would not have helped us because we proved

sin x
lim
= 1 in order to find the derivative of
x →0 x
sin x.

( ) ≤ lim

x 2 sin 1x
tan x

x2
.
x →0 tan x

x2
x 2 cos x
=
tan x
sin x
x 2 cos x
⎡⎛ x ⎞
⎤
= lim ⎢⎜
⎟ x cos x ⎥ = 0 .
x →0 sin x
x →0 ⎣⎝ sin x ⎠
⎦
lim

Thus, lim

x 2 sin

( 1x ) = 0 .

x →0 tan x
A table of values or graphing utility confirms
this.

27. a.

OB = cos t , BC = sin t and AB = 1 – cos t , so the area of triangle ABC is

The area of the sector COA is
region ABC is

1
sin t (1 – cos t ).
2

1
1
t while the area of triangle COB is cos t sin t , thus the area of the curved
2
2

1
(t – cos t sin t ).
2

1 sin t (1 – cos t )
area of triangle ABC
= lim 2
1
t →0+ area of curved region ABC t →0+ 2 (t – cos t sin t )

lim

sin t (1 – cos t )
cos t – cos 2 t + sin 2 t
4sin t cos t – sin t
4 cos t – 1 3
= lim
= lim
= lim
=
+ t – cos t sin t
+ 1 – cos 2 t + sin 2 t
+
+
4 cos t sin t
4 cos t
4
t →0
t →0
t →0
t →0
(L’Hôpital’s Rule was applied twice.)
= lim

1
1
1
t cos 2 t , so the area of the curved region BCD is cos t sin t – t cos 2 t.
2
2
2
1 cos t (sin t – t cos t )
area of curved region BCD
= lim 2
lim
1 (t – cos t sin t )
+ area of curved region ABC
t →0
t →0+
2

b. The area of the sector BOD is

cos t (sin t – t cos t )
sin t (2t cos t – sin t )
2t (cos 2 t – sin 2 t )
t (cos 2 t – sin 2 t )
= lim
= lim
= lim
2
2
t – sin t cos t

4 cos t sin t
2 cos t sin t
t →0+
t →0+ 1 – cos t + sin t
t →0+
t →0+

= lim

cos 2 t – 4t cos t sin t – sin 2 t

1– 0 – 0 1
=
2–0
2
2 cos t – 2sin t
t →0
(L’Hôpital’s Rule was applied three times.)
= lim

+

478

Section 8.1

2

2

=

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28. a.

Note that ∠DOE has measure t radians. Thus the coordinates of E are (cost, sint).
Also, slope BC = slope CE . Thus,
0− y
sin t − 0
=
(1 − t ) − 0 cos t − (1 − t )
(1 − t ) sin t
cos t + t − 1
(t − 1) sin t
y=
cos t + t –1
(t – 1) sin t
lim y = lim
+
+ cos t + t – 1
t →0
t →0
0
This limit is of the form .
0

(t – 1) sin t
sin t + (t – 1) cos t 0 + (–1)(1)
=
= –1
lim
= lim
– sin t + 1
–0 + 1
t →0+ cos t + t – 1 t →0+
−y =

b. Slope AF = slope EF . Thus,
t
t − sin t
=
1 − x 1 − cos t
t (1 − cos t )
= 1− x
t − sin t
t (1 + cos t )
x = 1−
t − sin t
t cos t – sin t
x=
t – sin t
t cos t – sin t
lim x = lim
+
+
t – sin t

t →0
t →0
0
The limit is of the form . (Apply l’Hôpital’s Rule three times.)
0
t cos t – sin t
–t sin t
= lim
lim
+
+
t – sin t
t →0
t →0 1 – cos t
– sin t – t cos t
t sin t – 2 cos t 0 – 2
= lim
= lim
=
= –2
+
+
sin t
cos t
1
t →0
t →0
ex −1
ex
⎛0⎞

29. By l’Hộpital’s Rule ⎜ ⎟ , we have lim f ( x) = lim
= lim
= 1 and
x
⎝0⎠
x →0 +
x →0+
x →0+ 1
ex −1
ex
= lim
= 1 so we define f (0) = 1 .
x →0− x
x →0− 1

lim f ( x) = lim

x →0 −

1
ln x
⎛0⎞
30. By l’Hộpital’s Rule ⎜ ⎟ , we have lim f ( x) = lim
= lim x = 1 and
⎝0⎠
x →1+
x →1+ x − 1 x →1+ 1
1
ln x
= lim x = 1 so we define f (1) = 1 .

lim f ( x) = lim
x →1−
x →1− x − 1 x →1− 1

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31. A should approach 4πb 2 , the surface area of a sphere of radius b.
2 2
⎡
2πa 2 b arcsin a a– b
⎢
2
lim ⎢ 2πb +
a →b + ⎢
a 2 – b2
⎣
Focusing on the limit, we have

lim

a →b

a 2 – b2
a
2

a 2 arcsin

+

2

2 2
⎤
a 2 arcsin a a– b
⎥
2
⎥ = 2πb + 2πb lim+
a →b
a 2 – b2
⎥
⎦

2a arcsin
= lim
a →b

a –b

+

a 2 – b2

a

⎛
+ a2 ⎜
⎝a

a
a 2 –b2

b
2

a –b

2

⎞
⎟
2
2
⎛
⎞
⎠ = lim ⎜ 2 a 2 – b 2 arcsin a – b + b ⎟ = b.
⎟
a
a →b + ⎜
⎝
⎠

Thus, lim A = 2πb 2 + 2πb(b) = 4πb 2 .

a →b +

32. In order for l’Hôpital’s Rule to be of any use, a(1)4 + b(1)3 + 1 = 0, so b = –1 – a.
Using l’Hôpital’s Rule,
ax 4 + bx3 + 1
4ax3 + 3bx 2
lim
= lim
x →1 ( x – 1) sin πx
x →1 sin πx + π( x – 1) cos πx
To use l’Hôpital’s Rule here,
4a(1)3 + 3b(1)2 = 0, so 4a + 3b = 0, hence a = 3, b = –4.
36 x 2 – 24 x
12
6
3 x 4 – 4 x3 + 1
12 x3 – 12 x 2
= lim
=
=–
= lim
2
–2π
π
x →1 2π cos πx – π ( x – 1) sin πx
x →1 ( x – 1) sin πx
x →1 sin πx + π( x – 1) cos πx
lim

a = 3, b = –4, c = –

6
π

33. If f ′(a ) and g ′(a ) both exist, then f and g are
both continuous at a. Thus, lim f ( x) = 0 = f (a )

38.

x →a

and lim g ( x ) = 0 = g (a ).
x →a

lim

x→a

f ( x)
f ( x) – f (a )
= lim
g ( x) x→a g ( x) – g (a )

f ( x )– f ( a )
x–a
lim
x → a g ( x )– g ( a )
x–a

=

cos x – 1 +

x2
2

34. lim

x →0

35. lim

x

36. lim

x →0

4

ex – 1 – x –

x →0

f ( x )– f ( a )
x–a
x →a
g ( x )– g ( a )
lim
x–a

x→a

lim

x

x2
2

=

–

4

1 – cos( x 2 )
3

x sin x

=

=

f ′(a)
g ′(a )

1
24
x3

6

=

1
24

1
2

tan x − x
sec2 x − 1
= lim 1
=2
x → 0 arcsin x − x
x →0
−1
2

37. lim

1− x

480

Section 8.1

The slopes are approximately 0.02 / 0.01 = 2 and
0.01/ 0.01 = 1 . The ratio of the slopes is
therefore 2 /1 = 2 , indicating that the limit of the

ratio should be about 2. An application of
l'Hopital's Rule confirms this.

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


41.

39.

The slopes are approximately 0.005 / 0.01 = 1/ 2
and 0.01/ 0.01 = 1 . The ratio of the slopes is
therefore 1/ 2 , indicating that the limit of the
ratio should be about 1/ 2 . An application of
l'Hopital's Rule confirms this.

The slopes are approximately 0.01/ 0.01 = 1 and
−0.01/ 0.01 = 1 . The ratio of the slopes is
therefore −1/1 = −1 , indicating that the limit of
the ratio should be about −1 . An application of
l'Hopital's Rule confirms this.
42. If f and g are locally linear at zero, then, since
lim f ( x ) = lim g ( x ) = 0 , f ( x) ≈ px and

40.

x →0

x →0

g ( x) ≈ qx , where p = f '(0) and q = g '(0) .

Then f ( x) / g ( x) ≈ px / px = p / q when x is
near 0.

The slopes are approximately 0.01/ 0.01 = 1 and
0.02 / 0.01 = 2 . The ratio of the slopes is
therefore 1/ 2 , indicating that the limit of the
ratio should be about 1/ 2 . An application of
l'Hopital's Rule confirms this.

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8.2 Concepts Review
1.

f ′( x)
g ′( x )

2. lim

x →a

∞
.
∞
3sec x tan x

5. The limit is of the form
lim

x→ π
2

f ( x)
g ( x)
or lim
1
x →a 1
g ( x)
f ( x)

3sec x + 5
= lim
tan x
x→ π

= lim

x→ π
2

3. ∞ – ∞, 0°, ∞°, 1∞

3 tan x
= lim 3sin x = 3
sec x x→ π
2

ln sin 2 x
= lim
x →0+ 3ln tan x x →0+
lim

x →0+

1. The limit is of the form

∞
.
∞

7. The limit is of the form

1 1000 x999
1000
ln x1000
= lim x
lim

x
1
x →∞
x →∞
1000
= lim
=0
x →∞ x

2. The limit is of the form

∞
. (Apply l’Hôpital’s
∞

Rule twice.)
2x

x →∞

x →∞

= lim

x →∞

x ⋅ 2 ln 2(1 + x ln 2)
x

x

x →∞

2

x →∞ 2 x

e

x

( 1x )

ln 2(1 + x ln 2)
=0

∞
4. The limit is of the form . (Apply l’Hôpital’s
∞
Rule three times.)
3x
3
= lim
lim =
1
x →∞ ln(100 x + e x ) x →∞
(100 + e x )
x
100 x + e

= lim

x →∞

= lim

x →∞

100 + e x

3e x
ex

=3

= lim

x →∞

)

=0

8. The limit is of the form

–∞
. (Apply l’Hôpital’s
∞

ln(4 – 8 x) 2

lim
= lim
–
tan πx
x→ 1
x→ 1

( 2)

1
(4–8 x )2

( 2)

= lim

(2)

–
x→ 1

–

2(4 – 8 x)(–8)

π sec2 πx

–16 cos 2 πx
32π cos πx sin πx
= lim

–
π(4 – 8 x )
–8π
x→ 1

(2)

300 + 3e x
ex

(2)

–
x→ 1

∞
.
∞
cot x
– csc2 x
= lim
1
– ln x x →0+ –

9. The limit is of the form

lim

x →0 +

2 x – ln x

= lim

2 x – ln x

sin 2 x
⎡ 2x
⎤
= lim ⎢
csc x – ln x ⎥ = ∞
+ ⎣ sin x
⎦
x →0
x
since lim
= 1 while lim csc x = ∞ and
+ sin x
x →0 +
x →0
x →0

lim

Section 8.2

x ln x1000

x →∞

x →0 +

482

1000

= lim

(

1
1 1000 x999
ln x1000 x1000
1
x

= lim – 4 cos πx sin πx = 0

= 0 (See Example 2).

300 x + 3e x

ln(ln x1000 )
= lim
lim
ln x
x →∞
x →∞

∞

.
∞

Rule twice.)

2 x ln 2

= lim

x ⋅ 2 x ln 2
2

10000

lim

2(ln x) 1x

x →∞

2 ln x

= lim

3.

= lim

1 2sin x cos x
sin 2 x

3 sec 2 x
tan x

2 cos 2 x 2
=
3
3

= lim

Problem Set 8.2

(ln x)2

–∞
.
–∞

6. The limit is of the form

4. ln x

lim

sec 2 x

2

+

– ln x = ∞.

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


10. The limit is of the form

∞
, but the fraction can
∞

15. The limit is of the form 00.
2

Let y = (3 x) x , then ln y = x 2 ln 3x

be simplified.
2 csc2 x
2
2
= lim
= =2
lim
x →0 cot 2 x
x →0 cos 2 x 12
1000

11. lim ( x ln x
x →0

The limit is of the form
lim

= lim

1
x

x →0

x →0 +

x

–

x →0

x →0

1
x2

2

13. lim (csc 2 x – cot 2 x) = lim
x →0

14.

x →0 +

lim csc x(ln(cos x)) = lim

x →0

The limit is of the form

ln(cos x)
sin x

0
.
0

1 (– sin x )
ln(cos x)
= lim cos x
cos x
x →0 sin x
x →0
sin x
0
= lim –
=– =0
1
x →0 cos 2 x

lim

sin 2 x

=1

lim (cos x)csc x = lim eln y = 1

x

x →0

lim (tan x – sec x) = lim

x→ π
2

x→ π
2

sin x – 1
cos x

2

x →0

17. The limit is of the form 0∞ , which is not an

indeterminate form.

0
The limit is of the form .
0
sin x – 1
cos x
0
= lim
=
=0
lim
–1
x → π cos x
x → π – sin x

lim (5cos x) tan x = 0
–
x →(π / 2 )

2

2
2
⎛ x 2 – sin 2 x ⎞
⎛
1 ⎞
⎛ 1
1 ⎞
18. lim ⎜ csc2 x –

= lim ⎜
=
lim
–
⎟
⎟
⎜
⎟
x →0 ⎜ x 2 sin 2 x ⎟
x →0 ⎝
x →0 ⎝ sin 2 x x 2 ⎠
x2 ⎠
⎝
⎠

Consider lim

x 2 – sin 2 x
2

x →0

2

lim

+

x →0

1 – cos 2 x

x →0

x →0 sin 2

x →0 +

x2
=0
2

Let y = (cos x)csc x , then ln y = csc x(ln(cos x))

⎛ x ⎞
12. lim 3 x 2 csc 2 x = lim 3 ⎜
⎟ = 3 since
x →0
x →0 ⎝ sin x ⎠
x
lim
=1
x →0 sin x

sin x

= lim –

16. The limit is of the form 1∞.

x →0

= lim

1
x2

1 ⋅3
3x
2
x →0 + – 3
x

= lim

∞
.
∞

lim (3x) x = lim eln y = 1

= lim – 1000 x = 0

2

1
x2

+

2

1000 x999

1

x →0

ln 3 x

lim

∞
.
∞

1000

+

The limit is of the form

1
x

x →0

ln x1000

x →0

ln x1000

) = lim

ln 3 x

lim x 2 ln 3 x = lim

x sin x

2

x – sin x

x →0

= lim

2

x sin x

x →0 sin 2

= lim

2

. The limit is of the form

2

2 x – 2sin x cos x

= lim

x →0 2 x sin

2

2

x + 2 x sin x cos x
2

2

2

2

x + 4 x sin x cos x + x cos x – x sin x

x →0 12 cos

0
. (Apply l’Hôpital’s Rule four times.)
0

= lim

1 – cos 2 x + sin 2 x

2

x →0

x – sin x cos x
2

x sin x + x 2 sin x cos x

4sin x cos x

= lim

x →0 6 x cos x 2

+ 6 cos x sin x − 4 x 2 cos x sin x − 6 x sin 2 x

4 cos 2 x – 4sin 2 x
2

2

2

2

2

2

x – 4 x cos x – 32 x cos x sin x – 12sin x + 4 x sin x

=

4 1
=
12 3

2

2
⎛ x 2 – sin 2 x ⎞
1
⎛1⎞
=⎜ ⎟ =
Thus, lim ⎜
⎟
2
2
⎜
⎟
9
x →0 x sin x
⎝ 3⎠
⎝
⎠

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19. The limit is of the form 1∞.

24. The limit is of the form 1∞.

Let y = ( x + e x / 3 )3 / x , then ln y =

3
ln( x + e x / 3 ).
x

3
3ln( x + e x / 3 )
ln( x + e x / 3 ) = lim
x
x →0 x
x →0
0
The limit is of the form .
0

(

3 + ex / 3

= lim

x →0

x+e

x/3

=

x →0 x 2

)

x →0

20. The limit is of the form (–1)0 .
The limit does not exist.
21. The limit is of the form 10 , which is not an
indeterminate form.
lim (sin x)cos x = 1

x2

x →0

0
.
0
(Apply l’Hôpital’s rule twice.)
1 (– sin x )
ln(cos x)
− tan x
lim
= lim cos x
= lim
2
2x
x →0
x →0
x →0 2 x
x
− sec 2 x −1
1
=
=−
2
2
2
x →0
2

lim (cos x)1/ x = lim eln y = e−1/ 2 =
x →0

1
e

25. The limit is of the form 0∞ , which is not an
indeterminate form.
lim (tan x) 2 / x = 0

x →0 +

26. The limit is of the form ∞ + ∞, which is not an
indeterminate form.

x→ π
2

∞

22. The limit is of the form ∞ , which is not an
indeterminate form.
lim x x = ∞

lim (e – x – x) = lim (e x + x) = ∞

x→ – ∞

x →∞

27. The limit is of the form 00. Let
y = (sin x) x , then ln y = x ln(sin x).

x →∞

23. The limit is of the form ∞ 0 . Let
1
y = x1/ x , then ln y = ln x.
x
1
ln x
lim ln x = lim
x →∞ x
x →∞ x
–∞
.
The limit is of the form
∞
1

ln x
1
= lim x = lim = 0
lim
x →∞ x
x →∞ 1
x →∞ x
lim x

ln(cos x) .

ln(cos x)

ln(cos x ) = lim

x →0

x →∞

x2

= lim

lim ( x + e x / 3 )3 / x = lim eln y = e 4

1/ x

1

The limit is of the form

4
=4
1

x →0

1

lim

lim

3
1 + 13 e x / 3
3ln( x + e x / 3 )
x +e x / 3
lim
= lim
x
1
x →0
x →0

2

Let y = (cos x)1/ x , then ln y =

= lim e
x →∞

ln y

=1

ln(sin x)

lim x ln(sin x) = lim

x →0

+

x →0

1
x

+

–∞
.
∞
1 cos x
sin x

The limit is of the form
lim

x →0

ln(sin x)

+

1
x

= lim

x →0 +

–

1
x2

⎡ x
⎤
= lim ⎢
(– x cos x) ⎥ = 1 ⋅ 0 = 0
+ ⎣ sin x
⎦
x →0
lim (sin x ) x = lim eln y = 1

x →0 +

x →0+

28. The limit is of the form 1∞. Let

1
ln(cos x – sin x).
x
1
ln(cos x − sin x)
lim ln(cos x − sin x ) = lim
x
x →0 x
x →0
y = (cos x – sin x)1/ x , then ln y =

= lim

1
(− sin x − cos x)
cos x −sin x

1
− sin x − cos x
= lim
= −1
x →0 cos x − sin x
x →0

lim (cos x − sin x )1/ x = lim eln y = e−1

x →0

484

Section 8.2

x →0

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


29. The limit is of the form ∞ – ∞.

1⎞
1⎞
x – sin x
⎛
⎛ 1
lim ⎜ csc x – ⎟ = lim ⎜
– ⎟ = lim
x ⎠ x →0 ⎝ sin x x ⎠ x →0 x sin x
x →0 ⎝
0
The limit is of the form . (Apply l’Hôpital’s
0
Rule twice.)
x – sin x
1 – cos x
= lim
lim
x →0 x sin x
x →0 sin x + x cos x
sin x
0
= lim
= =0
2
x →0 2 cos x – x sin x

x

⎛ 1⎞
⎛ 1⎞

Let y = ⎜ 1 + ⎟ , then ln y = x ln ⎜ 1 + ⎟ .
x
⎝
⎠
⎝ x⎠

(

ln 1 + 1x
⎛ 1⎞
lim x ln ⎜ 1 + ⎟ = lim
1
x →∞
⎝ x ⎠ x →∞
x

lim

(

ln 1 + 1x

x →∞

1
x

) = lim

x →∞

)

0
.
0

1
1+ 1

x

(– )

–

1
x2

1
x2

1
=1
x →∞ 1 + 1
x

= lim

x

31. The limit is of the form 3∞ , which is not an
indeterminate form.
lim (1 + 2e )

x →0 +

=∞

32. The limit is of the form ∞ – ∞.
x ⎞
ln x – x 2 + x
⎛ 1
lim ⎜
–
⎟ = lim
x →1 ⎝ x – 1 ln x ⎠ x →1 ( x – 1) ln x
0
.
0
Apply l’Hôpital’s Rule twice.
1 − 2x +1
ln x − x 2 + x
lim
= lim x
x →1 ( x − 1) ln x
x →1 ln x + x −1
x

The limit is of the form

2

1− 2x + x
−4 x + 1 −3
3
= lim
=
=−
2
2
x →1 x ln x + x − 1 x →1 ln x + 2

= lim

1
ln(cos x).
x

1
ln(cos x)
ln(cos x) = lim
x
x
x →0
0
The limit is of the form .
0
lim

x →0

1

(– sin x)
ln(cos x)
sin x
= lim cos x
= lim –
=0
1
x
x →0
x →0
x →0 cos x
lim

lim (cos x)1/ x = lim eln y = 1
x →0

34. The limit is of the form 0 ⋅ – ∞.
ln x
lim ( x1/ 2 ln x) = lim
x →0 +

The limit is of the form
lim

x →0 +

1
x

x →0+

ln x
1
x

= lim

x →0 +

–

–∞
.
∞

1
x
1
2 x3/ 2

= lim – 2 x = 0
x →0+

35. Since cos x oscillates between –1 and 1 as
x → ∞, this limit is not of an indeterminate form
previously seen.

Let y = ecos x , then ln y = (cos x)ln e = cos x

⎛ 1⎞
lim ⎜1 + ⎟ = lim eln y = e1 = e
x⎠
x →∞ ⎝
x →∞

x 1/ x

Let y = (cos x)1/ x , then ln y =

x →0

30. The limit is of the form 1∞.

The limit is of the form

33. The limit is of the form 1∞.

Instructor's Resource Manual

lim cos x does not exist, so lim ecos x does not

x →∞

x →∞

exist.

36. The limit is of the form ∞ – ∞.
lim [ln( x + 1) – ln( x – 1)] = lim ln

x →∞

x →∞

x +1
x –1

1 + 1x

x +1
x +1
= lim
= 1, so lim ln
=0
1
x –1
x →∞ x – 1 x →∞ 1 –
x →∞
lim

x

37. The limit is of the form

0
, which is not an
–∞

indeterminate form.
x
lim
=0
x →0+ ln x
38. The limit is of the form – ∞ ⋅ ∞, which is not an
indeterminate form.
lim (ln x cot x) = – ∞
x →0 +

Section 8.2

485

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


1 + e−t > 1 for all t, so

39.

d.

1 + e−t dt > ∫ dt = x − 1 .

x

x

∫1

lim n

n →∞

1 + e−t dt

x

∫1

lim

x →∞

1+ e
1

= lim

x →∞

x

−x

x

lim

∫1

+

x →1

sin t dt

x −1

= lim

+

x →1

n

since lim

n →∞

=1

n

lim

1
n

= lim

sin x
= sin(1)
1

n

n→∞

42. a.

n

n −1

= lim

1
n

n →∞

lim

a = lim e

ln y

n →∞

x →0 +

n →∞

n →∞

lim

n

n →∞

= lim

n →∞

a −1
1
n
n

−

n→∞

a ln a = ln a

Section 8.2

1
x

1
x

1
x
1
x →0 + – 2
x
ln y

= lim

x →0 +

–∞
.
∞

= lim – x = 0
x →0 +

=1

b. The limit is of the form 10 , since

lim x x = 1 by part a.

x →0 +

lim x ln( x x ) = 0

x →0+

lim ( x x ) x = lim eln y = 1

n

1
n

1 n
n2

−

a ln a
1
n2

x →0 +

Note that 10 is not an indeterminate form.

a −1

a = 1 by part a.

= lim

ln x

Let y = ( x x ) x , then ln y = x ln( x x ).

0
This limit is of the form ,
0
since lim

ln x

x →0 +

( n a − 1) = nlim
→∞

+

lim x x = lim e

n = lim eln y = 1

n

x →0

x →0 +

n →∞

lim n

1
n2

n (ln n − 1) = ∞

+

lim

1

n →∞

−

The limit is of the form

ln n
= lim n = 0
n →∞ n
n→∞ 1
n

1

n2

The limit is of the form 00.

x →0

lim

lim

( ) (1 − ln n)

n →∞

lim x ln x = lim

=1

b. The limit is of the form ∞ 0 .
1
Let y = n n , then ln y = ln n .
n
1
ln n
lim ln n = lim
n →∞ n
n →∞ n
∞
.
This limit is of the form

∞

n

Let y = x x , then ln y = x ln x.

1
Let y = a , then ln y = ln a.
n
1
lim ln a = 0
n →∞ n
n

0
,
0

n = 1 by part b.

n

n →∞

486

n −1

This limit is of the form

0
40. This limit is of the form .
0

c.

n

1

∞
.
The limit is of the form
∞

41. a.

( n n − 1) = nlim
→∞

c.

The limit is of the form 01 , since
lim x x = 1 by part a.

x →0 +

x

Let y = x( x ) , then ln y = x x ln x

lim x x ln x = – ∞

x →0+

lim x( x

x →0

+

x

)

= lim eln y = 0
x →0 +

Note that 01 is not an indeterminate form.

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


d. The limit is of the form 10 , since

1

ln x

= lim x = 0, so lim x1/ x = lim eln y = 1
x →∞ x
x →∞ 1
x →∞
x →∞
lim

lim ( x x ) x = 1 by part b.

x →0 +

1 ln x

Let y = (( x ) ) , then ln y = x ln(( x ) ).
x x x

x x

lim x ln(( x x ) x ) = 0

x →0 +

lim (( x x ) x ) x = lim eln y = 1

x →0 +

x →0 +

Note that 10 is not an indeterminate form.
e.

The limit is of the form 00 , since
lim ( x

(xx )

x →0 +

Let y = x( x

) = 0 by part c.

( xx )

)

x →0

x →0

x →0 +

ln x
1
x
x( x )

+

x

x( x )

x

2

x

)

Note: lim x(ln x )2 = lim
x →0 +

lim x( x

x →0 +

( xx )

)

x

(ln x)2

x →0 +

2 ln x
x
1

x →0+ – 2
x

x →0 +

c.

x
x →0+ – x ( x x ) ⎡ x x (ln x +1) ln x + x ⎤
⎢
x ⎥⎦
⎣

= lim

lim (1x + 2 x )1/ x = ∞

lim (1x + 2 x )1/ x = 0

1
x

– x( x

The limit is of the form (1 + 1)∞ = 2∞ , which
is not an indeterminate form.

x →0 –

x x(ln x) + x x ln x + x

0
=
=0
1⋅ 0 + 1⋅ 0 + 1
x →0

44. a.

–∞
.
∞

= lim

x

y ′ < 0 on (e, ∞). When x = e, y = e1/ e .

b. The limit is of the form (1 + 1) – ∞ = 2 – ∞ ,
which is not an indeterminate form.

1

+

x
( x( x ) )2

= lim

y is maximum at x = e since y ′ > 0 on (0, e) and

, then ln y = x( x ) ln x.

The limit is of the form
lim

⎛ 1 ln x ⎞ 1x ln x
y′ = ⎜
−
⎟e
⎝ x2 x2 ⎠
y ′ = 0 when x = e.

x

x
ln x
lim x( x ) ln x = lim
+

y = x1/ x = e x

1
x

= lim – 2 x ln x = 0
x →0 +

= lim eln y = 1

x →0+

The limit is of the form ∞0 .
Let y = (1x + 2 x )1/ x , then
ln y =

1
ln(1x + 2 x )
x

1
ln(1x + 2 x )
ln(1x + 2 x ) = lim
x
x →∞ x
x →∞
∞
The limit is of the form . (Apply
∞
l’Hôpital’s Rule twice.)
1 (1x ln1 + 2 x ln 2)
ln(1x + 2 x )
1x + 2 x
lim
= lim
1
x
x →∞
x →∞
lim

= lim

2 x ln 2

x →∞ 1x

+ 2x

2 x (ln 2)2

= lim

x →∞ 1x

ln1 + 2 x ln 2

= ln 2

lim (1x + 2 x )1/ x = lim eln y = eln 2 = 2

x →∞

x →∞

d. The limit is of the form 10 , since 1x = 1 for
all x. This is not an indeterminate form.

43.

lim (1x + 2 x )1/ x = 1

x →−∞

ln x
x
ln x
= −∞, so lim x1/ x = lim eln y = 0
lim
+ x
x →0
x →0+
x →0+

ln y =

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Section 8.2

487

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


45.

1/ t

1k + 2k + " + n k

lim

c.

n k +1

n →∞

1 ⎡⎛ 1 ⎞ ⎛ 2 ⎞
⎛n⎞
⎢⎜ ⎟ + ⎜ ⎟ + " ⎜ ⎟
n →∞ n ⎢⎝ n ⎠
⎝n⎠
⎝n⎠
⎣
k

k

= lim

k

⎤
⎥
⎥⎦

9 ⎞

⎛1
lim ⎜ 2t + 5t ⎟
+ ⎝ 10
10 ⎠
t →0

= 10 2 ⋅

10 9

5 ≈ 4.562

48. a.

k

n
1 ⎛i⎞
= lim ∑ ⋅ ⎜ ⎟
n →∞
i =1 n ⎝ n ⎠
The summation has the form of a Reimann sum
for f ( x ) = x k on the interval [ 0,1] using a

regular partition and evaluating the function at
1
i
each right endpoint. Thus, Δxi = , xi = , and
n
n

b.

k

⎛i⎞
f ( xi ) = ⎜ ⎟ . Therefore,
⎝n⎠
1k + 2k + " + n k

lim

n k +1

n →∞

e
2

1 ⎛i⎞
= lim ∑ ⋅ ⎜ ⎟
n →∞
i =1 n ⎝ n ⎠
n

lim

k

1
⎡ 1 k +1 ⎤

= ∫ x k dx = ⎢
x ⎥
0
⎣ k +1
⎦0
1
=
k +1

1/ t

⎞
1 ⎛ n
, then ln y = ln ⎜ ∑ ci xit ⎟ .
⎜
⎟
t ⎝ i =1
⎠

⎛ n
⎞
ln ⎜ ∑ ci xit ⎟
⎜
⎟
⎞
1 ⎛ n
⎠
lim ln ⎜ ∑ ci xit ⎟ = lim ⎝ i =1
⎜
⎟

+ t
+
t
t →0
⎝ i =1
⎠ t →0

The limit is of the form

0
, since
0

⎛
⎞
ln ⎜ ∑ ci xit ⎟
⎜
⎟
⎠ = lim
lim ⎝ i =1
t
t →0 +
t →0+

n

∑ ci = 1.
i =1

n

∑ ci xit ln xi

n

∑ ci xit i =1
i =1

n

n

i =1

i =1
1/ t

= ei =1

= lim eln y
t →0 +

n

= x1c1 x2c2 … xncn = ∏ xi ci
i =1

1/ t

47. a.

1 ⎞
⎛1
lim ⎜ 2t + 5t ⎟
+⎝2
2 ⎠
t →0

b.

4 ⎞
⎛1
lim ⎜ 2t + 5t ⎟
+⎝5
5
⎠
t →0

1/ t

488

Section 8.2

= lim

n →∞

lim

n →∞

c.

1

∫0 xe

2nx

xenx

xe
−x

= lim

2x

n →∞ x 2 enx

nx

= 2 5 ≈ 3.162
5

= 5 2 ⋅ 54 ≈ 4.163

∞
.

∞
=0

1
2
dx = ⎡ − xe− x − e− x ⎤ = 1 −
⎣
⎦0
e

1

−2 x

1
3
dx = ⎡ −2 xe−2 x − e−2 x ⎤ = 1 −
⎣
⎦0
e2

1

−3 x

1
4
dx = ⎡ −3xe−3 x − e−3 x ⎤ = 1 −
⎣
⎦0

e3

∫0 4 xe
∫0 9 xe

1
5
dx = ⎡ −4 xe−4 x − e−4 x ⎤ = 1 −
⎣
⎦0
e4

1

−4 x

1

−5 x

∫016 xe

∫0 25 xe
∫0 36e

∞
.
∞

2nx

−6 x

1
6
= ⎡ −5 xe−5 x − e−5 x ⎤ = 1 −
⎣
⎦0
e5

1
7
dx = ⎡ −6 xe−6 x − e−6 x ⎤ = 1 −
⎣
⎦0
e6

d. Guess: lim

∫

1 2

n →∞ 0

n xe− nx dx = 1
1

1 2

n

∑ ln xici

, so the limit is of the form

xe− nx dx = ⎡ − nxe− nx − e− nx ⎤
⎣
⎦0
n +1
= −(n + 1)e− n + 1 = 1 −
en
1
⎛ n +1⎞
lim ∫ n 2 xe− nx dx = lim ⎜1 −
⎟
0
n →∞
n →∞ ⎝
en ⎠
n +1
= 1 − lim
if this last limit exists. The
n →∞ e n
∞
.
limit is of the form
∞
n +1
1

lim
= lim
= 0, so
n →∞ e n
n →∞ en

∫0 n

= ∑ ci ln xi = ∑ ln xi ci
⎛ n
⎞
lim ⎜ ∑ ci xi t ⎟
⎜
⎟
t →0+ ⎝ i =1
⎠

nx

This limit is of the form

1

n

1

n x

n →∞ e nx

1

⎛ n
⎞
46. Let y = ⎜ ∑ ci xit ⎟
⎜
⎟
⎝ i =1
⎠

n2 x

n 2 xe− nx =

1 2

∫n
n →∞ 0
lim

xe− nx dx = 1 .

Instructor's Resource Manual

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49. Note f(x) > 0 on [0, ∞).
⎛ x 25 x3 ⎛ 2 ⎞ x ⎞

lim f ( x) = lim ⎜
+
+⎜ ⎟ ⎟ = 0
x →∞
x →∞ ⎜ e x
e x ⎝ e ⎠ ⎟⎠
⎝
Therefore there is no absolute minimum.
f ′( x) = (25 x 24 + 3 x 2 + 2 x ln 2)e− x
− ( x 25 + x3 + 2 x )e− x

7.

8.

= (− x 25 + 25 x 24 − x3 + 3 x 2 − 2 x + 2 x ln 2)e− x
Solve for x when f ′( x) = 0 . Using a numerical
method, x ≈ 25.
A graph using a computer algebra system verifies
that an absolute maximum occurs at about x = 25.

8.3 Concepts Review
1. converge

3.

∫– ∞ f ( x)dx; ∫0

∞

0

f ( x)dx

11.

In this section and the chapter review, it is understood
means lim [ g ( x)]
b →∞

b
a

∞

x

∞

3.

∫1

∞

dx

∞

= ⎡ 1 + x 2 ⎤ = ∞ – 82 = ∞

∫9
⎢⎣
⎥⎦ 9
2
1+ x
The integral diverges.
x dx

∞

⎡ x⎤
2
∫1 πx = ⎢2 π ⎥ = ∞ – π = ∞
⎣
⎦1
The integral diverges.
∞

dx

x

1
dx = [ln(ln x)]e∞ = ∞ – 0 = ∞
x ln x
The integral diverges.
∞

∫e

∞

1
⎡1
2⎤
∫e x dx = ⎢⎣ 2 (ln x) ⎥⎦ e = ∞ – 2 = ∞
The integral diverges.
∞ ln x

b

b

ln 2 + 1
⎡ ln x 1 ⎤
= lim ⎢ −
− ⎥ =
b →∞ ⎣
x
x ⎦2
2

14.

∞

∫1

xe – x dx

u = x, du = dx

1

1
1
⎡1
⎤
4. ∫ e4 x dx = ⎢ e4 x ⎥ = e4 – 0 = e4
–∞
4
⎣4
⎦ –∞ 4

6.

∞

⎡
⎤
1
∫1 (1 + x2 )2 dx = ⎢⎢ – 2(1 + x2 ) ⎥⎥
⎣
⎦1
∞

b 1
⎡ ln x ⎤
= lim ⎢ −
+ lim

dx
b →∞ ⎣
x ⎥⎦ 2 b →∞ ∫2 x 2

∞

2
2
1
2 xe – x dx = ⎡⎢ – e – x ⎤⎥ = 0 – (– e –1 ) =
e
⎣
⎦1

1

∞

∞

dx

–5

⎡ 1 ⎤
1
1
∫– ∞ x 4 = ⎢⎣ – 3x3 ⎥⎦ = – 3(–125) – 0 = 375
–∞
5

∞

)⎤
⎦10

1
1
1
dx, dv =
dx, v = − .
2
x
x
x
∞ ln x
b ln x
dx
∫2 x 2 dx = blim
→∞ ∫2 x 2

= ∞ – e100 = ∞
dx = ⎡ e x ⎤
⎣ ⎦100
The integral diverges.

∫100 e

2

13. Let u = ln x, du =

2.

5.

12.

and likewise for

similar expressions.
1.

1

⎛ 1⎞ 1
= 0–⎜– ⎟ =
⎝ 4⎠ 4

Problem Set 8.3
∞
a

x

⎡ x 0.00001 ⎤
9. ∫
=⎢
⎥ = ∞ – 100, 000 = ∞
1 x 0.99999

⎣⎢ 0.00001 ⎦⎥1
The integral diverges.

4. p > 1

that [ g ( x)]

∞

∫10 1 + x2 dx = 2 ⎡⎣ln(1 + x

∞

b

∫ cos x dx
b →∞ 0
lim

dx

1
= ∞ – ln 101 = ∞
2
The integral diverges.

10.

2.

∞

⎡
1
⎤
∫1 x1.00001 = ⎢⎣ – 0.00001x0.00001 ⎥⎦
1
1
1
⎛
⎞
= 0–⎜–
= 100, 000
⎟=
⎝ 0.00001 ⎠ 0.00001
∞

Instructor’s Resource Manual

dv = e – x dx, v = – e – x
∞

∫1

∞

∞

xe – x d = ⎡ – xe – x ⎤ + ∫ e – x dx
⎣

⎦1
1

∞
2
= ⎡ – xe – x – e – x ⎤ = 0 – 0 – (– e –1 – e –1 ) =
⎣
⎦1
e
1

⎡
⎤
1
= ⎢–
15. ∫
⎥
3
2
– ∞ (2 x – 3)
⎣⎢ 4(2 x – 3) ⎦⎥ – ∞
1

dx

=–

1
1
– (–0) = −

4
4

Section 8.3

489

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


16.

∞

1/ 3 ⎤ ∞

dx

∫4 (π − x )2 / 3 = ⎡⎣−3 (π − x )

⎦4

= ∞ + 33 π − 4 = ∞

The integral diverges.
17.

∞

x

∫– ∞

x2 + 9

dx = ∫

0

x

–∞

x2 + 9

dx + ∫

0

The integral diverges since both

18.

∞

0

dx

0

1

dx

–1

2

x +9

∞

dx

∞

0

dx = ⎡ x 2 + 9 ⎤ + ⎡ x 2 + 9 ⎤ = (3 – ∞) + (∞ – 3)
2
⎣⎢
⎦⎥ – ∞ ⎢⎣
⎦⎥ 0
x +9
x

x

∫– ∞

∫– ∞ ( x2 + 16)2 = ∫– ∞ ( x2 + 16)2 + ∫0
∫ ( x 2 + 16)2 = 128 tan

∞

dx and

∞

∫0

x
2

x +9

dx diverge.

dx
2

( x + 16) 2

x
x
by using the substitution x = 4 tan θ.
+
2

4 32( x + 16)
0

⎡ 1
⎤
x
⎡ 1 ⎛ π⎞ ⎤
π
–1 x
∫– ∞ ( x2 + 16)2 = ⎢⎢128 tan 4 + 32( x2 + 16) ⎥⎥ = 0 – ⎢⎣128 ⎜⎝ – 2 ⎟⎠ + 0⎥⎦ = 256
⎣
⎦ –∞
0

dx

∞

⎡ 1
⎤
x
1 ⎛ π⎞
π
–1 x
∫0 ( x2 + 16)2 = ⎢⎢128 tan 4 + 32( x2 + 16) ⎥⎥ = 128 ⎜⎝ 2 ⎟⎠ + 0 – (0) = 256
⎣
⎦0
∞
dx
π

π
π
∫– ∞ ( x 2 + 16)2 = 256 + 256 = 128
∞

19.

dx

1

∞

1

∞

0

1

1

1

∫ ( x + 1)2 + 9 dx = 3 tan

–1

0

1
⎡1
–1 1 1 ⎛ π ⎞
–1 x + 1 ⎤
∫– ∞ ( x + 1)2 + 9 dx = ⎢⎣ 3 tan 3 ⎥⎦ – ∞ = 3 tan 3 – 3 ⎜⎝ – 2 ⎟⎠ =
∞

∫0

1⎛
–1 1 ⎞
⎜ π + 2 tan
⎟
6⎝
3⎠

∞

1⎛

1

∞

∞

∫– ∞
For

dx

x + 1⎤
1⎛ π⎞ 1
1 1⎛
1⎞
⎡1
= ⎜ ⎟ – tan –1 = ⎜ π – 2 tan –1 ⎟
dx = ⎢ tan –1
⎥
2
3 6⎝
3⎠
3 ⎦0 3 ⎝ 2 ⎠ 3
⎣3
( x + 1) + 9
1

∫– ∞ x2 + 2 x + 10 dx = 6 ⎜⎝ π + 2 tan
20.

( x + 1)2 + 9

x +1
by using the substitution x + 1 = 3 tan θ.
3

1

0

1

∞

∫– ∞ x 2 + 2 x + 10 dx = ∫– ∞ ( x + 1)2 + 9 dx = ∫– ∞ ( x + 1)2 + 9 dx + ∫0

x
e

2x
0

dx = ∫

0

x

– ∞ e –2 x

x

∞

x

0

2x

dx + ∫

0

∫– ∞ e –2 x dx = ∫– ∞ xe

2x

e

1⎛
–1 1 ⎞ π
⎟ + ⎜ π – 2 tan
⎟=
3⎠ 6 ⎝
3⎠ 3

–1 1 ⎞

dx

dx, use u = x, du = dx, dv = e2 x dx, v =

0

1 2x
e .
2

0

1 0 2x
1
1
⎡ 1 2x ⎤
⎡ 1 2x 1 2x ⎤
2x
∫– ∞ xe dx = ⎢⎣ 2 xe ⎥⎦ – ∞ – 2 ∫– ∞ e dx = ⎢⎣ 2 xe – 4 e ⎥⎦ – ∞ = 0 – 4 – (0) = – 4
∞ x
∞
1
For ∫
dx = ∫ xe –2 x dx, use u = x, du = dx, dv = e –2 x dx, v = – e –2 x .
0 e2 x
0
2
0

∞

∞

1 ∞
1
1⎞ 1
⎡ 1
⎤
⎡ 1
⎤

⎛
xe –2 x dx = ⎢ – xe –2 x ⎥ + ∫ e –2 x dx = ⎢ – xe –2 x – e –2 x ⎥ = 0 – ⎜ 0 – ⎟ =
0
4
4⎠ 4
⎣ 2
⎦0 2
⎣ 2
⎦0
⎝
∞ x
1 1
∫– ∞ 2 x dx = – 4 + 4 = 0
e
∞

∫0

490

Section 8.3

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


21.

∞

25. The area is given by
∞
∞⎛ 1
2
1 ⎞
∫1 4 x 2 − 1dx = ∫1 ⎜⎝ 2 x –1 – 2 x + 1 ⎟⎠ dx

∞

0

∫– ∞ sech x dx = ∫– ∞ sech x dx = ∫0 sech x dx
= [tan –1 (sinh x)]0– ∞ + [tan –1 (sinh x)]∞
0

∞

⎡ ⎛ π ⎞⎤ ⎡ π
⎤
= ⎢0 – ⎜ – ⎟ ⎥ + ⎢ – 0 ⎥ = π
2
2
⎝
⎠
⎣
⎦
⎣
⎦

22.

∞

csch x dx = ∫

∫1

∞

1

∞

2e

x

1

2x

–1

=∫

e

=

1⎛
⎛ 1 ⎞⎞ 1
= ⎜ 0 − ln ⎜ ⎟ ⎟ = ln 3
2⎝
⎝ 3 ⎠⎠ 2
2x −1
Note:. lim ln =
= 0 since
2x + 1
x →∞
⎛ 2x −1 ⎞
.
lim ⎜
⎟ =1
x →∞ ⎝ 2 x + 1 ⎠

∞
1
2
dx = ∫
dx
1 ex – e– x
sinh x

dx

Let u = e x , du = e x dx .
2e x

∞

∫1

e

2x

–1

dx = ∫

∞

e

∞⎛ 1
1 ⎞
du = ∫ ⎜
–
⎟ du
e
⎝ u –1 u + 1 ⎠
u –1

2

26. The area is
∞
∞⎛ 1

1
1 ⎞
∫1 x 2 + x dx = ∫1 ⎜⎝ x – x + 1 ⎟⎠ dx

2

∞

⎡ u –1 ⎤
= [ln(u –1) – ln(u + 1)]∞
e = ⎢ln
⎥
⎣ u + 1⎦ e
e –1
= 0 – ln
≈ 0.7719
e +1
b –1
b –1 ⎞
⎛
= 0 since lim
= 1⎟
⎜ lim ln
b
b +1 ⎠
+
1
b
→∞
b

→∞
⎝

23.

24.

∞

x ⎤
1
∞ ⎡
= ⎡⎣ln x − ln x + 1 ⎤⎦ = ⎢ ln
= 0 − ln = ln 2
⎥
1
2
⎣ x + 1 ⎦1
.

27. The integral would take the form
∞ 1
∞
k∫
dx = [ k ln x ]3960 = ∞
3960 x
which would make it impossible to send anything
out of the earth's gravitational field.

∞

⎡ 1
⎤
cos x dx = ⎢
(sin x − cos x) ⎥
∫
x
⎣ 2e
⎦0
1
1
= 0 − (0 − 1) =
2
2
(Use Formula 68 with a = –1 and b = 1.)
∞ −x
e
0

28. At x = 1080 mi, F = 165, so
k = 165(1080) 2 ≈ 1.925 × 108 . So the work done
in mi-lb is
∞
∞
1
1.925 × 108 ∫
dx = 1.925 × 108 ⎡ − x −1 ⎤
⎣
⎦1080
1080 x 2

8
1.925 × 10
=
≈ 1.782 × 105 mi-lb.
1080

∞

⎡ 1
⎤
= ⎢−
(cos x + sin x) ⎥
∫
x
⎣ 2e
⎦0
1
1
= 0 + (1 + 0) =
2
2
(Use Formula 67 with a = –1 and b = 1.)
∞ −x
e sin x dx
0

1
1 ⎡ 2x −1 ⎤
∞
⎡ ln 2 x − 1 − ln 2 x + 1 ⎤⎦ = ⎢ ln

1
2⎣
2 ⎣ 2 x + 1 ⎥⎦1

∞

∞

0

0

29. FP = ∫ e− rt f (t ) dt = ∫ 100, 000e−0.08t
∞

⎡ 1
⎤
= ⎢−
100, 000e−0.08t ⎥ = 1,250,000
⎣ 0.08
⎦0
The present value is $1,250,000.

∞

30. FP = ∫ e−0.08t (100, 000 + 1000t )dt
0

∞

= ⎡ −1, 250, 000e−0.08t − 12,500te−0.08t − 156, 250e−0.08t ⎤ = 1,406,250
⎣
⎦0
The present value is $1,406,250.

31.

a.

∞

a

b

1

∞

∫−∞ f ( x) dx = ∫−∞ 0 dx + ∫a b − a dx + ∫b
= 0+

0 dx

1
1
(b − a )
[ x ]b + 0 =
b−a a
b−a

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491

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


b.

μ=∫

∞

−∞

=∫

a

x f ( x) dx
b

−∞

x ⋅ 0 dx + ∫ x
a

∞
1
dx + ∫ x ⋅ 0 dx
b
b−a

b

1 ⎡ x2 ⎤
= 0+
⎢ ⎥ +0
b − a ⎢⎣ 2 ⎥⎦
a

2

2

=

b −a
2(b − a)

=

(b + a)(b − a )
2(b − a)

=

a+b
2

σ2 = ∫

∞

−∞

=∫

a

( x − μ ) 2 dx
b

−∞

= 0+

( x − μ )2 ⋅ 0 dx + ∫ ( x − μ )2
a

3 ⎤b

1 ⎡( x − μ )
⎢
⎥ +0
b−a ⎢

3
⎥
⎣
⎦a
3

=

∞
1
dx + ∫ ( x − μ )2 ⋅ 0 dx
b
b−a

3

1 (b − μ ) − ( a − μ )
b−a
3

1 b3 − 3b 2 μ + 3bμ 2 − a3 + 3a 2 μ − 3a μ 2
b−a
3
Next, substitute μ = (a + b) / 2 to obtain
=

σ2 =
=

=

c.

1
( b − a )3
12 ( b − a )

( b − a )2
12

0

−∞

=

a.

2

P ( X < 2) = ∫
=∫

32.

1
⎡ 1 b3 − 3 b 2 a + 3 ba 2 − 1 a3 ⎤
4
4
4
⎦

3(b − a ) ⎣ 4

−∞
2

0 dx + ∫

0

f ( x) dx

1
dx
10 − 0

2 1
=
10 5

∞

x
θ (θ )

∞β

0

∫−∞ f ( x) dx = ∫−∞ 0 dx + ∫0

β −1 −( x / θ ) β

e

dx

In the second integral, let u = ( x / θ ) β . Then,
du = ( β / θ )(t / θ ) β −1 dt . When x = 0, u = 0 and when
x → ∞, u → ∞ . Thus,
∞

∞β

∫−∞ f ( x) dx = ∫0
∞

(x)

θ θ

β −1 − ( x / θ ) β

e

dx

∞

= ∫ e−u du = ⎡ −e−u ⎤ = −0 + e0 = 1
⎣

⎦0
0

492

Section 8.3

Instructor’s Resource Manual

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


b.

μ=∫

∞

−∞

xf ( x) dx = ∫

0

−∞

x ⋅ 0 dx + ∫

∞

0

β
θ

⎛ x⎞
x⎜ ⎟
⎝θ ⎠

β −1

e−( x / θ ) dx∂

2 ∞ 2 − ( x / 3)2
3
π
x e
dx =
3 ∫0
2

=

σ2 = ∫

∞

−∞

( x − μ )2 f ( x) dx = ∫

0

−∞

( x − μ )2 ⋅ 0 dx +

2
2 ∞
( x − μ )2 xe− ( x / 9) dx
∫
0
9

3
3
3
π −μ =
π − π =0
2
2
2
The probability of being less than 2 is
=

c.

2

f ( x ) dx = ∫

∫−∞

0

2β

( )

0 dx + ∫ θ θx
−∞
0

β

β −1 − ( x / θ ) β

e

2

β
dx = 0 + ⎡⎢ −e −( x / θ ) ⎤⎥
⎣
⎦0

2

= 1 − e−(2 / θ ) = 1 − e−(2 / 3) ≈ 0.359

33.

f ′( x) = –

x–μ

σ

3

2π

2
2
e –( x – μ ) / 2σ

2

2
2
( x – μ ) –( x – μ )2 / 2σ 2
e –( x – μ ) / 2σ +
e
3
σ 2π
σ 5 2π

1

f ′′( x) = –

⎛ ( x – μ )2
1 ⎞ –( x – μ )2 / 2σ 2
–
=⎜
=
⎟e
⎜ σ 5 2π σ 3 2π ⎟
⎝
⎠
2
2
1
[( x − μ )2 − σ 2 ]e –( x – μ ) / 2σ
σ 5 2π
f ′′( x) = 0 when ( x – μ )2 = σ 2 so x = μ ± σ and the distance from μ to each inflection point is σ.

34.

a.

b.

⎡ 1
dx = CM k ⎢ –
M x k +1
⎣ kx k

∞

f ( x)dx = ∫

∫– ∞
μ=∫

∞

–∞

∞

CM k

∞

kM k

M

k +1

xf ( x)dx = ∫ x

dx = kM k ∫

∞

1 ⎞ C
C

⎤
k⎛
= . Thus, = 1 when C = k.
⎥ = CM ⎜ 0 +
k ⎟
k
kM ⎠ k
⎦M
⎝

∞

M

b 1
⎛
⎞
dx = kM k ⎜ lim ∫
dx ⎟
k
M
x
x
⎝ b→∞
⎠

1

k

x
This integral converges when k > 1.
b
⎛
⎡
⎤
1
k⎜
When k > 1, μ = kM
lim ⎢ –
⎥
⎜⎜ b→∞ ⎢ (k –1) x k –1 ⎥
⎣
⎦
M
⎝

⎞
⎛
⎞ kM
1
⎟ = kM k ⎜ –0 +
⎟=
k –1 ⎟ k –1
⎜
⎟⎟
(k –1) M
⎝
⎠
⎠

The mean is finite only when k > 1.

Instructor’s Resource Manual

Section 8.3

493

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


c.

Since the mean is finite only when k > 1, the variance is only defined when k > 1.
2
⎛ 2 2kM
∞
∞⎛
kM ⎞ kM k
k 2M 2 ⎞ 1
k ∞
=
+
kM
x
–
x
dx

σ 2 = ∫ ( x – μ ) 2 f ( x)dx = ∫ ⎜ x –
dx
⎜
⎟
⎟
∫M ⎜
–∞
M⎝
k –1
k –1 ⎠ x k +1
(k –1)2 ⎟⎠ x k +1
⎝
2k 2 M k +1 ∞ 1
k 3M k +2 ∞ 1
dx +
dx
∫
∫
M x k –1
M xk
k –1
(k –1) 2 M x k +1
The first integral converges only when k – 1 > 1 or k > 2. The second integral converges only when k > 1,
which is taken care of by requiring k > 2.
= kM k ∫

1

∞

dx –

∞

∞

⎤
⎤
2k 2 M k +1 ⎡
1
k 3M k +2
+
–
–
⎥
⎢
⎥
k –1 ⎣⎢ (k –1) x k –1 ⎦⎥
(k –1)2
⎢⎣ (k – 2) x k –2 ⎦⎥ M
M

⎡

1

σ 2 = kM k ⎢ –

∞

⎡ 1 ⎤
⎢– k ⎥
⎣ kx ⎦ M

⎛
⎞ 2k 2 M k +1 ⎛
⎞ k 3M k +2 ⎛
1
1
1 ⎞
= kM k ⎜ –0 +
+
–
–0 +
–0 +
⎟
⎜
⎟
⎟
k –2 ⎟
k –1 ⎟
2 ⎜
⎜
⎜
k
–1
(k – 2) M
(k –1) M
kM k ⎠
⎝

⎠
⎝
⎠ (k –1) ⎝
=

kM 2 2k 2 M 2 k 2 M 2
+
–
k – 2 (k –1) 2 (k –1) 2

⎛ k 2 – 2k + 1 – k 2 + 2k ⎞
⎛ 1
kM 2
k ⎞
= kM 2 ⎜
= kM 2 ⎜
=
–
⎟
⎟
⎜ k – 2 (k –1)2 ⎟
⎜ (k – 2)(k –1) 2
⎟ (k – 2)(k –1)2
⎝
⎠
⎝
⎠

35. We use the results from problem 34:
a.

To have a probability density function (34 a.)
we need C = k ; so C = 3. Also,
kM
μ=
(34 b.) and since, in our problem,
k −1
μ = 20, 000 and
k =3, we have
20000 =

3
4 × 104
M or M =
.
2
3

b. By 34 c., σ 2 =

kM 2
(k − 2)(k − 1)

4 ⎞2

2

so that

36. u = Ar ∫

c.

∞

∫105

37. a.

Thus 6

25
$100,000.

494

sin x dx
0

a →−∞

Both do not converge since –cos x is
oscillating between –1 and 1, so the integral
diverges.
b.

a

[− cos x]− a
∫ sin x dx = alim

a →∞ − a
→∞
lim

a

= lim [− cos a + cos(−a)]
a →∞

= lim [− cos a + cos a] = lim 0 = 0
a →∞

38. a.

of one percent earn over

Section 8.3

∞

0

a

3

⎛ 4 × 10
1⎤
64
⎡ 1

=⎜
−
=
⎟ lim
⎜ 3 ⎟ t →∞ ⎣⎢1015 t 3 ⎦⎥ 27 × 103
⎝
⎠
≈ 0.0024

∞

∫−∞ sin x dx = ∫−∞ sin x dx + ∫0
a →∞

t
⎞
⎡1⎤
⎟ lim ⎢ 3 ⎥
⎟ t →∞ ⎣ x ⎦ 5
10
⎠

4 ⎞3

∞

= lim [ − cos x ]0 + lim [ − cos x ]a

3

⎛ 4 × 104
−⎜
⎜ 3
⎝

( r + x 2 )3 / 2

⎤
⎞
A⎛
a
⎟
⎥ = ⎜1 −
⎟
r ⎝⎜
⎥⎦ a
r 2 + a2 ⎠
dx
x
=
Note that ∫
by using
2
2 3/ 2
(r + x )
r 2 r 2 + x2
the substitution x = r tan θ .

8

⎛ 4 × 104 ⎞
t
3
f ( x) dx = ⎜
dx =
⎟ lim
⎜ 3 ⎟ t →∞ ∫105 x 4
⎝
⎠

dx
2

A⎡
x
= ⎢
r ⎣⎢ r 2 + x 2

3 ⎛ 4 × 10
4 × 10
⎟ =
⎟
4⎝ 3 ⎠
3

σ2 = ⎜
⎜

∞

a

b.

a→∞

The total mass of the wire is
∞ 1
π
∫0 1 + x2 dx = 2 from Example 4.
∞

⎡1
2 ⎤
∫0 1 + x2 dx = ⎢⎣ 2 ln 1 + x ⎥⎦0 which
diverges. Thus, the wire does not have a
center of mass.
∞

x

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


39. For example, the region under the curve y =

⎡

⎤
1 ⎤
1
⎡
, n + 1⎥
⎢ n, n + 2 ⎥ and ⎢ n + 1 –
2
2n ⎦
2(n + 1)
⎣
⎢⎣
⎥⎦
1
1
will never overlap since
≤ and
2
2
2n
1
1
≤ .
2
8
2(n + 1)

1
x

to the right of x = 1.

Rotated about the x-axis the volume is
∞ 1
π∫
dx = π . Rotated about the y-axis, the
1 x2
∞
1
volume is 2π ∫ x ⋅ dx which diverges.
1
x
40. a.

The graph of f consists of a series of isosceles
triangles, each of height 1, vertices at
1
1
⎛
⎞
⎛
⎞
⎜ n – 2 , 0 ⎟ , (n, 1), and ⎜ n + 2 , 0 ⎟ ,
2n
2n
⎝
⎠
⎝
⎠
based on the x-axis, and centered over each
integer n.
lim f ( x) does not exist, since f(x) will be 1

Suppose lim f ( x) = M ≠ 0, so the limit
x →∞

exists but is non-zero. Since lim f ( x) = M ,
x →∞

there is some N > 0 such that when x ≥ N,
M
f ( x) – M ≤
, or
2
M
M
M–
≤ f ( x) ≤ M +
2
2
Since f(x) is nonnegative, M > 0, thus
M
> 0 and
2
∞

∫0

f ( x )dx = ∫

N

0

f ( x)dx + ∫

∞
N

x →∞

at each integer, but 0 between the triangles.
Each triangle has area
1
1⎡
1
1 ⎞⎤
⎛
bh = ⎢ n +
–⎜n –
⎟ ⎥ (1)
2
2
2⎣
2n
2n 2 ⎠ ⎦
⎝
=

f ( x)dx
∞

M
⎡ Mx ⎤
dx = ∫ f ( x)dx + ⎢
⎥ =∞
N 2
0
0
⎣ 2 ⎦N
so the integral diverges. Thus, if the limit
exists, it must be 0.
≥∫

b.

N

f ( x)dx + ∫

∞

N

For example, let f(x) be given by
1
⎧ 2
3
⎪2n x – 2n + 1 if n – 2 ≤ x ≤ n
2n
⎪
1

⎪
f ( x) = ⎨ –2n 2 x + 2n3 + 1 if n < x ≤ n +
2n 2
⎪
⎪0
otherwise
⎪
⎩
for every positive integer n.
⎛
⎛
1 ⎞
1 ⎞
3
f ⎜n –
= 2n 2 ⎜ n –
⎟ – 2n + 1
2⎟
2n ⎠
2n 2 ⎠
⎝
⎝
3

3

= 2n – 1 – 2n + 1 = 0
f ( n ) = 2 n 2 ( n ) – 2 n3 + 1 = 1

1⎛ 1 ⎞

1
⎜ ⎟=
2 ⎝ n 2 ⎠ 2n 2

∞

∫0

f ( x)dx is the area in all of the triangles,

thus
∞

∫0

∞

f ( x)dx = ∑

1

n =1 2n

2

=

1 ∞ 1
∑
2 n =1 n 2

=

1 1 ∞ 1 1 1 ∞ 1
+ ∑
≤ +
dx
2 2 n = 2 n 2 2 2 ∫1 x 2

=

1 1 ⎡ 1⎤
1 1
+ –
= + (–0 + 1) = 1
2 2 ⎢⎣ x ⎥⎦1
2 2

∞

∞

(By viewing

1

∑ n2

as a lower Riemann sum

n=2

for

1
x2

Thus,

)
∞

∫0

f ( x )dx converges, although

lim f ( x) does not exist.

x →∞

lim f (n) = lim (–2n 2 x + 2n3 + 1) = 1 = f (n)

x→n+

x →n+

⎛
⎛
1 ⎞
1 ⎞

3
f ⎜n+
= –2n 2 ⎜ n +
⎟ + 2n + 1
2⎟
2n ⎠
2n 2 ⎠
⎝
⎝

= –2n3 –1 + 2n3 + 1 = 0
Thus, f is continuous at
1
1
n–
, n, and n +
.
2
2n
2n 2
Note that the intervals

Instructor’s Resource Manual

Section 8.3

495

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

41.

∫1

1.1

x

∫1

x

100 1

x

1
⎡
⎤
dx = ⎢ –
0.01 ⎥
⎣ 0.01x
⎦1

100

⎡ x 0.01 ⎤
dx = ⎢

⎥
x 0.99
⎣⎢ 0.01 ⎦⎥1

∫1

1

10

∫0

2

dx =

≈ 4.71

1 ⎡ −1 ⎤10
tan x
⎦0
π⎣

π(1 + x )
1.4711
≈
≈ 0.468
π
50
1

1
−1 50
∫0 π(1 + x2 ) dx = π ⎡⎣ tan x ⎤⎦0
1.5508
≈
≈ 0.494
π
100
1
1
−1 100
∫0 π(1 + x2 ) dx = π ⎡⎣ tan x ⎤⎦ 0
1.5608
≈
≈ 0.497
π
1

1

2

2π
1

∫0
∫0

2π
3 1

∫0

2π
4 1

∫0

2π

dx

33 2
3(b – 1) 2 / 3
3
3
=
2 – lim
–0=
3
3
2
2
2
2
b→1+
3

⎡
⎤

3
= lim ⎢ –
2. ∫
⎥
1/ 3
1 ( x – 1) 4 / 3 b →1+
⎢⎣ ( x – 1) ⎥⎦ b
3
3
3
=–
+ lim
=–
+∞
3
1/ 3
3
+
2 b→1 ( x –1)
2
The integral diverges.

3.

3

dx

10

dx

∫3

10

= lim ⎡ 2 x – 3 ⎤⎦
b
x – 3 b→3+ ⎣

= 2 7 – lim 2 b – 3 = 2 7
b →3+

4.

9

∫0

dx

b

= lim ⎡ −2 9 – x ⎤⎦
0
9 – x b→9 – ⎣

= lim − 2 9 – b + 2 9 = 6
b →9 –

5.

exp(–0.5 x 2 )dx ≈ 0.3413
exp(–0.5 x 2 )dx ≈ 0.4772

3

⎡ 3( x – 1) 2 / 3 ⎤
⎢
⎥
∫1 ( x – 1)1/ 3 = blim
2
→1+ ⎣⎢
⎦⎥ b
3

=

≈ 4.50

dx = [ln x]100
1 = ln100 ≈ 4.61
1

100

≈ 3.69

1.

100

1.01

∫1

43.

1 ⎤
⎡
dx = ⎢ –
⎥
⎣ 0.1x 0.1 ⎦1

1

100

Problem Set 8.4

= 0.99
100

1

100

∫1

42.

100

⎡ 1⎤
dx = ⎢ – ⎥
2
⎣ x ⎦1
x
1

100

6.

1

b

dx

= lim ⎡sin –1 x ⎤
⎣
⎦0
2
b →1–
1– x
π
π
= lim sin –1 b – sin –1 0 = – 0 =
–

2
2
b →1

∫0

∞

b

dx = lim ⎡ 1 + x 2 ⎤
∫100
⎥⎦100
2
b →∞ ⎢⎣
1+ x
x

exp(–0.5 x 2 )dx ≈ 0.4987

= lim 1 + b 2 + 10, 001 = ∞

exp(–0.5 x 2 )dx ≈ 0.5000

The integral diverges.

b →∞

7.

3

1

b

b

8.4 Concepts Review
1. unbounded
2. 2
3.

lim

b→4

∫

b

– 0

1
4– x

4. p < 1

496

Section 8.4

dx

1

3

1

∫–1 x3 dx = blim
∫ 3 dx + blim
∫ 3 dx
→0 – –1 x
→0+ b x
3

⎡ 1 ⎤
⎡ 1 ⎤
= lim ⎢ –
+ lim ⎢ –
2⎥
2⎥
–
+
b →0 ⎣ 2 x ⎦ –1 b →0 ⎣ 2 x ⎦ b

⎛
1
1⎞ ⎛ 1

1 ⎞
= ⎜ lim –
+ + – + lim
2 2 ⎟ ⎜ 18
2⎟
–
+
b →0 2b ⎠
⎝ b→0 2b
⎠ ⎝
1⎞ ⎛ 1
⎛
⎞
= ⎜ −∞ + ⎟ + ⎜ – + ∞ ⎟
2⎠ ⎝ 8
⎝
⎠
The integral diverges.

Instructor's Resource Manual

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


1

–5

∫5

8.

x

2/3

1

b

1

–5

∫ 2 / 3 dx + blim
∫
b →0 + 5 x
→0– b

dx = lim

x

2/3

9.

dx

= lim

= lim ⎡3x1/ 3 ⎤ + lim ⎡3 x1/ 3 ⎤
⎣
⎦ 5 b→0 – ⎣
⎦b
b →0+

b

dx
–5 / 7

128 –5 / 7

∫
b →0 + b

dx + lim

x

dx

128

b

⎡7
⎤

⎡7
⎤
= lim ⎢ x 2 / 7 ⎥ + lim ⎢ x 2 / 7 ⎥
– ⎣2
+
⎦ –1 b→0 ⎣ 2
⎦b
b →0
7
7
7
7
= lim b 2 / 7 – (–1)2 / 7 + (128) 2 / 7 – lim b 2 / 7
– 2
+ 2
2
2
b →0
b→0
7 7
21
= 0 – + (4) – 0 =
2 2
2

= lim 3b1/ 3 – 33 5 + 33 –5 – lim 3b1/ 3
b →0+
3

x

∫ x
b →0 – –1

–5

b

128 –5 / 7

∫–1

b →0 –
3

= 0 – 3 5 + 33 5 – 0 = 33 −5 − 3 5 = −6 3 5

10.

1

∫0 3

x
1 – x2

dx = lim

∫

x

b

b →1– 0 3

1 – x2

dx

b

⎡ 3
⎤
= lim ⎢ – (1 – x 2 )2 / 3 ⎥
–⎣ 4
⎦0
b →1
3
3
3 3
= lim − (1 – b 2 ) 2 / 3 + = –0 + =
– 4
4
4 4
b →1
4

dx

0

(2 – 3x)1/ 3

11. ∫

= lim
b→ 2

∫

– 0

3

dx

b

(2 – 3 x)1/ 3

+ lim
b→ 2

∫

4

+ b

3

4

b

dx
(2 – 3 x)1/ 3

⎡ 1
⎤
⎡ 1
⎤
= lim ⎢ – (2 – 3 x)2 / 3 ⎥ + lim ⎢ – (2 – 3x ) 2 / 3 ⎥
–
+
⎦ 0 b→ 2 ⎣ 2
⎦b
b→ 2 ⎣ 2
3

3

1
1
1
1
= lim − (2 – 3b) 2 / 3 + (2)2 / 3 – (–10) 2 / 3 + lim (2 – 3b)2 / 3
–
+

2
2
2
b→ 2
b→ 2 2
3

3

1
1
1
= 0 + 22 / 3 − 102 / 3 + 0 = (22 / 3 − 102 / 3 )
2
2
2

12.

13.

8

∫

5

x
2 2/3

(16 − 2 x )

–4

x

∫0

16 – 2 x

=

lim

2

dx =

⎡ 3
⎤
dx = lim ⎢ − (16 − 2 x 2 )1/ 3 ⎥
−⎣ 4
⎦
b→ 8
lim
b→ – 8

b

x

dx +
+ ∫0
16 – 2 x 2

b→ – 8

5

3
3
3
= lim − (16 − 2b 2 )1/ 3 + 3 6 = 3 6
− 4
4
4
b→ 8
–4

– ∫b

x
16 – 2 x 2

dx

–4

b

⎡ 1
⎤
⎡ 1
⎤
– ln 16 – 2 x 2 ⎥ + lim ⎢ – ln 16 – 2 x 2 ⎥
+⎢
–
4
4
⎣
⎦
⎣
⎦b
0 b→ – 8
8

b→ –

=

lim

b

1
1
1
1
lim − ln 16 – 2b 2 + ln16 – ln16 + lim
ln 16 – 2b 2

+ 4
–
4
4
4
b→ – 8
b→ – 8

1
⎡
⎤ ⎡ 1
⎤
= ⎢ –(– ∞) + ln16 ⎥ + ⎢ – ln16 + (– ∞) ⎥
4
⎣
⎦ ⎣ 4
⎦
The integral diverges.

14.

15.

3

∫0

–1

b

dx = lim ⎡ – 9 – x 2 ⎤ = lim − 9 – b 2 + 9 = 3
⎢
⎥⎦ 0 b→3–
2
b →3 – ⎣
9– x
x

dx

∫–2 ( x + 1)4 / 3

b

⎡
⎤
3
3
3
= lim –
+
= –(– ∞) – 3
= lim ⎢ –
⎥
1/
3
1/
3
–

(–1)1/ 3
b → –1– ⎣⎢ ( x + 1)
⎦⎥ –2 b→ –1 (b + 1)

The integral diverges.

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Section 8.4

497

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


16. Note that
3

dx

⎡

dx

1

1

⎤

∫ x2 + x − 2 = ∫ ( x − 1)( x + 2) = ∫ ⎢⎣ 3( x − 1) − 3( x + 2) ⎥⎦ dx

dx

3

dx

b

by using a partial fraction decomposition.

dx

+ lim ∫
∫0 x2 + x – 2 = blim
∫ 2
2
→1– 0 x + x – 2 b→1+ b x + x – 2
3

b

1
1
⎡1
⎤
⎡1
⎤

= lim ⎢ ln x –1 – ln x + 2 ⎥ + lim ⎢ ln x –1 – ln x + 2 ⎥
– ⎣3
+
3
3
⎦ 0 b→1 ⎣ 3
⎦b
b →1
3

b

⎡1
x –1 ⎤
⎡1
x –1 ⎤
1 b –1 1 1 1 2
1 b –1
= lim ⎢ ln
⎥ + lim+ ⎢ 3 ln x + 2 ⎥ = lim– 3 ln b + 2 – 3 ln 2 + 3 ln 5 – lim+ 3 ln b + 2
– 3
x
+
2
b →1 ⎣
b →1
⎦ 0 b→1 ⎣
⎦ b b→1
1 1⎞ ⎛1 2
⎛

⎞
= ⎜ – ∞ – ln ⎟ + ⎜ ln + ∞ ⎟
3 2⎠ ⎝3 5
⎝
⎠
The integral diverges.
1

17. Note that

3

1

=

2

2

−

1
1
+
4( x − 1) 4( x + 1)

x − x − x + 1 2( x − 1)
3
b

3
dx
dx
dx
+ lim ∫
∫0 x3 – x2 – x + 1 = blim
– ∫0 x3 – x 2 – x + 1
+ b x3 – x 2 – x + 1
b→1
→1
3

b

⎡
⎤
⎡
⎤
1
1
1
1
1
1
= lim ⎢ –
– ln x − 1 + ln x + 1 ⎥ + lim ⎢ –
– ln x − 1 + ln x + 1 ⎥
–
+
4

4
b →1 ⎣ 2( x –1) 4
⎦ 0 b→1 ⎣ 2( x –1) 4
⎦b

⎡⎛
⎡ 1 1
⎛
1
1 b +1 ⎞ ⎛ 1
1
1 b + 1 ⎞⎤
⎞⎤
lim ⎢⎜ –
+ ln
+ ⎜ – + 0 ⎟ ⎥ + lim ⎢ – + ln 2 – ⎜ –
+ ln
⎟
⎟⎥
b −1 ⎠ ⎝ 2
⎠ ⎦ b→1+ ⎣ 4 4
b →1– ⎣⎝ 2(b –1) 4
⎝ 2(b –1) 4 b − 1 ⎠ ⎦
1⎞ ⎛ 1 1
⎛
⎞
= ⎜ ∞ + ∞ – ⎟ + ⎜ – + ln 2 + ∞ – ∞ ⎟
2⎠ ⎝ 4 4
⎝
⎠

The integral diverges.
x1/ 3

18. Note that
x

2/3

−9

1

=

1/ 3

x

+

9
1/ 3

x

( x 2 / 3 − 9)

x1/ 3

.

b

27
⎡ 3 2 / 3 27
⎤
⎛ 3 2 / 3 27
⎞ ⎛
⎞
2/3
2/3
∫0 x2 / 3 – 9 dx = b→lim27 – ⎢⎣ 2 x + 2 ln x – 9 ⎥⎦0 = b→lim27 – ⎜⎝ 2 b + 2 ln b – 9 ⎟⎠ – ⎜⎝ 0 + 2 ln 9 ⎟⎠
27
27
=
– ∞ – ln 9
2
2
The integral diverges.
27

19.

π/4

∫0

b

⎡ 1
⎤

tan 2 xdx = lim ⎢ – ln cos 2 x ⎥
–
2
⎦0
b→ π ⎣
4

1
1
= lim − ln cos 2b + ln1 = –(–∞) + 0
–
2
2
b→ π
4

The integral diverges.

20.

π/2

∫0

π/2

csc xdx = lim ⎡⎣ln csc x – cot x ⎤⎦
b
+
b →0

= ln 1 – 0 – lim ln csc b – cot b
b →0 +

= 0 – lim ln
b →0

+

1 – cos b
sin b

1 – cos b
0
is of the form .
0
b →0+ sin b
1 – cos b
sin b 0
= lim
= =0
lim
+ sin b
+ cos b
1
b →0
b →0
1 – cos b
Thus, lim ln
= – ∞ and the integral

+
sin b
b →0
diverges.
lim

498

Section 8.4

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


21.

π/2

∫0

1 − cos x
x
= sin 2 ,
2
2
1
1
x

= − csc 2 .
cos x − 1
2
2

sin x
π/ 2
dx = lim ⎡⎣ ln 1 – cos x ⎤⎦
b
+
1 – cos x
b →0

25. Since

= ln1 – lim ln 1 – cos b = 0 – (– ∞)
b →0+

The integral diverges.
22.

23.

π/2

⎡3 2/3 ⎤
sin
x⎥
∫0 3 sin x dx = blim
+⎢

⎦b
→0 ⎣ 2
3 2/3 3 2/3 3
= (1)
– (0)
=
2
2
2
π/2

π/2

∫0

cos x

π
b
– lim cot = 0 – ∞
2 b →0 +
2
The integral diverges.
= cot

b

⎡1
⎤
tan x sec x dx = lim ⎢ tan 3 x ⎥

–
3
⎦0
b→ π ⎣
2

π

dx
x⎤
⎡
cot ⎥
∫0 cos x – 1 = blim
+⎢
2 ⎦b
→0 ⎣
π

2

26.

–1

∫–3 x

dx

b

= lim ⎡ 2 ln(– x) ⎤⎦
–3
ln(– x) b→ –1– ⎣

= lim 2 ln(–b) – 2 ln 3 = 0 – 2 ln 3

2

b →−1–

1
1
= lim tan 3 b – (0)3 = ∞
–
3
3
b→ π

= –2 ln 3

2

The integral diverges.
27.
24.

π/4

∫0

sec2 x

b

1 ⎤
⎡
dx = lim ⎢ –
2
–
tan x – 1 ⎥⎦ 0
(tan x – 1)
b→ π ⎣

ln 3

∫0

–
b→ π
4

= lim ⎡ 2 e x –1 ⎤
⎥⎦ b
+⎢
x
e –1 b→0 ⎣

= 2 3 – 1 – lim 2 eb – 1 = 2 2 – 0 = 2 2
b →0+

4

= lim −

ln 3

e x dx

1
1
+
= –(– ∞) – 1
tan b – 1 0 – 1

The integral diverges.

28. Note that
4

∫2
29.

e

4 x − x 2 = 4 − ( x 2 − 4 x + 4) = 22 − ( x − 2)2 . (by completing the square)

dx
4 x – x2

= lim

∫

dx

b

b→4– 2

dx

4 x – x2

b

x – 2⎤
π
π
⎡
–1 b – 2
– sin –1 0 = – 0 =
= lim ⎢sin –1
⎥ = lim– sin
–⎣
2
2
2
2
⎦ 2 b→4
b→ 4

[ln(ln x)]b = ln(ln e) – lim ln(ln b) = ln 1 – ln 0 = 0 + ∞
∫1 x ln x = blim
→1+
b →1+
e

The integral diverges.
10

⎡
1
⎤
1
1
1
= lim –
30. ∫
=–
+ lim
=–
+∞
99
99
1 x ln100 x b →1+ ⎢⎣ 99 ln 99 x ⎥⎦
+
99 ln 10 b→1 99 ln b
99 ln 99 10
b
The integral diverges.

10

31.

dx

4c

⎡
⎤
= lim ⎢ln x + x 2 − 4c 2 ⎥ = ln ⎡⎣ (4 + 2 3)c ⎤⎦ − lim ln b + b 2 − 4c 2
+⎣
2
2
⎦b
b → 2c +
b → 2c
x − 4c
= ln ⎡⎣ (4 + 2 3)c ⎤⎦ − ln 2c = ln(2 + 3)
4c

∫2c

dx

Instructor’s Resource Manual

Section 8.4

499

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


32.

2c

x dx

∫c

=∫

x 2 + xc – 2c 2

2c

x dx

( x + 2c )

c

2

– 94 c 2

=∫

2c

c

( x + 2c ) dx − c 2c
dx
∫
0
2
2
2
( x + 2c ) − 94 c2
( x + 2c ) − 94 c2

⎡
c
c
= lim ⎢ x 2 + xc – 2c 2 − ln x + + x 2 + xc – 2c 2
+
2
2
b →c ⎣

2c

⎤
⎥
⎦b

c 5c
⎡
c
c
⎤
= 4c 2 − ln
+ 4c 2 – lim ⎢ b 2 + bc – 2c 2 − ln b + + b 2 + bc – 2c 2 ⎥
+
2
2
2
2
b →c ⎣
⎦
c 9c ⎛
c 3c
⎞
c 9c c 3c
c
= 2c − ln – ⎜ 0 − ln
+ 0 ⎟ = 2c − ln + ln = 2c − ln 3
2
2
2
2
2
2 2 ⎝
2
2
⎠

1

33. For 0 < c < 1,
1

dv =

x

is continuous. Let u =

x (1 + x)

1
1
, du = –
dx .
1+ x
(1 + x) 2

dx, v = 2 x .
1

⎡2 x ⎤
1
1
1
2 2 c
xdx
2 c

xdx
xdx
∫c x (1 + x) dx = ⎢1 + x ⎥ + 2∫c (1 + x)2 = 2 – 1 + c + 2∫c (1 + x)2 = 1 – 1 + c + 2∫c (1 + x)2
⎣
⎦c
1

1

⎡ 2 c
1
1
xdx
xdx ⎤
+ 2∫
dx = lim ⎢1 –
⎥ = 1 – 0 + 2∫0
2
c
1+ c
c →0 ⎢⎣
x (1 + x)
(1 + x) 2
(1 + x) ⎥⎦
This last integral is a proper integral.

Thus, lim ∫

1

1

c →0 c

1

34. Let u =

1+ x

1

dv =

x

, du = –

1
2(1 + x)3 / 2

dx

dx, v = 2 x .
1

⎡ 2 x ⎤
1
1
2 1 2 c

x
x
=⎢
dx =
–
+∫
dx
For 0 < c < 1, ∫
⎥ + ∫c
3
/
2
c x(1 + x)
c
2
1+ c
(1 + x)3 / 2
(1 + x)
⎣ 1 + x ⎦c
1

Thus,

1

∫0

dx

dx

x(1 + x)

= lim ∫

1

c →0 c

⎡
⎤
1
1
x
2 c
x
= lim ⎢ 2 –
+∫
dx ⎥ = 2 – 0 + ∫
dx
3
/
2
c
0
1+ c
x(1 + x) c→0 ⎢⎣
(1 + x)
(1 + x)3 / 2
⎥⎦

dx

This is a proper integral.
35.

3

∫–3

x
9 – x2

dx = ∫

= – 9 + lim

+

b → –3

36.

500

0

x

–3

9 – x2

9 – b 2 – lim
b →3

0

b

dx = lim ⎡ – 9 – x 2 ⎤ + lim ⎡ – 9 – x 2 ⎤
⎢
⎥⎦ b b→3– ⎢⎣
⎥⎦ 0
0
2
b→ –3+ ⎣
9– x

dx + ∫

3

x

9 – b 2 + 9 = –3 + 0 – 0 + 3 = 0

–

0

b

⎡ 1
⎤
⎡ 1
⎤
dx = lim ⎢ − ln 9 − x 2 ⎥ + lim ⎢ − ln 9 − x 2 ⎥
∫−3 9 − x2
−3 9 − x 2
0 9 − x2
+⎣ 2
−
⎦ b b→3 ⎣ 2
⎦0
b →3
1
1
ln 9 − b 2 − lim ln 9 − b 2 + ln 3 = (− ln 3 − ∞) + (∞ + ln 3)
= − ln 3 + lim
+ 2
− 2
b →−3
b →3
The integral diverges.
3

x

dx = ∫

Section 8.4

0

x

dx + ∫

3

x

Instructor’s Resource Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.