Calculus one and several variables 10E Salas solutions manual ch11

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU

JWDD027-11 JWDD027-Salas-v1 December 2, 2006 16:49

588 SECTION 11.1

CHAPTER 11

SECTION 11.1

1. lub = 2; glb = 0 2. lub = 2; glb = 0

3. no lub; glb = 0 4. lub = 1, no glb

5. lub = 2; glb = −2 6. lub = 3; glb = −1

7. no lub; glb = 2 8. lub = 2; glb = −2

9. lub = 212 ; glb = 2 10. lub = 0; glb = −1

11. lub = 1; glb = 0.9 12. lub = 219 , glb = −2 1

9

13. lub = e; glb = 0 14. no lub, glb = 1

15. lub = 12 (−1 +

√5); glb = 1

2 (−1 −√

5) 16. no lub, no glb

17. no lub; no glb 18. no lub; no glb

19. no lub; no glb 20. lub = 0; no glb

21. glb S = 0, 0 ≤(

111

)3< 0 + 0.001 22. glb = 1; 1 ≤ 1 < 1 + 0.0001

23. glb S = 0, 0 ≤(

110

)2n−1

< 0 +(

110

)k (n >

k + 12

)

24. glb = 0; 0 ≤(

12

)n

< 0 +(

14

)k

for n > 2k

25. Let ε > 0. The condition m ≤ s is satisfied by all numbers s in S. All we have to show therefore is

that there is some number s in S such that

s < m + ε.

Suppose on the contrary that there is no such number in S. We then have

m + ε ≤ x for all x ∈ S.

This makes m + ε a lower bound for S. But this cannot be, for then m + ε is a lower bound for S that

is greater than m, and by assumption, m is the greatest lower bound.

26. (a) Let M = |a1| + · · · + |an|. Then for any i, |ai| < M , so S is bounded

(b) lub S = max {a1, a2, . . . , an} ∈ S

glb S = min {a1, a2, . . . , an} ∈ S



SECTION 11.1 589

27. Let c = lub S. Since b ∈ S, b ≤ c. Since b is an upper bound for S, c ≤ b. Thus, b = c.

28. S consists of a single element, equal to lub S.

29. (a) Suppose that K is an upper bound for S and k is a lower bound. Let t be any element of T . Then

t ∈ S which implies that k ≤ t ≤ K. Thus K is an upper bound for T and k is a lower bound, and

T is bounded.

(b) Let a = glb S. Then a ≤ t for all t ∈ T. Therefore, a ≤ glb T. Similarly, if b = lub S, then

t ≤ b for all t ∈ T, so lub T ≤ b. It now follows that glb S ≤ glb T ≤ lub T ≤ lub S.

30. (a) Let S = {r : r <√

2, r rational}.(b) Let T = {t : t < 2, t irrational}.

31. Let c be a positive number and let S = {c, 2c, 3c, . . .}. Choose any positive number M and consider the

positive number M/c. Since the set of positive integers is not bounded above, there exists a positive

integer k such that k ≥ M/c. This implies that kc ≥ M . Since kc ∈ S, it follows that S is not

bounded above.

32. (a) If S is a set of negative numbers, then 0 is an upper bound for S. It follows that α = lub S ≤ 0.

(b) If T is a set of positive numbers, then 0 is a lower bound for T . It follows that β = glb T ≥ 0.

33. (a) See Exercise 75 in Section 1.2.

(b) Suppose x20 > 2. Choose a positive integer n such that

2x0

n− 1

n2< x2

0 − 2.

Then,2x0

n− 1

n2< x2

0 − 2 =⇒ 2 < x20 −

2x0

n+

1n2

=(x0 − 1

n

)2

(c) If x20 < 2, then choose a positive integer n such that

2x0

n+

1n2

< 2 − x20.

Then

x20 +

2x0

n+

1n2

< 2 =⇒(x0 + 1

n

)2< 2

34. Assume that there are only finitely many primes, p1, p2, · · · , pn and let Q = p1 · p2 · · · pn + 1. Q has

a prime divisor p. But Q is not divisible by any of the pi, so p �= p1 for all i a contradiction.

35. (a) n = 5 : 2.48832; n = 10 : 2.59374; n = 100 : 2.70481; n = 1000 : 2.71692; n = 10, 000 : 2.71815

(b) lub = e; glb = 2



590 SECTION 11.2

36. a2 a3 a4 a5 a6 a7 a8 a9 a10

94

53

65

12 −3 4 9

453

65

(b) a20 = 94 , a30 = −3, a40 = 6

5 , a50 = 94

(c) lubS = 4, glbS = −3

37. (a) a1 a2 a3 a4 a5

1.4142 1.6818 1.8340 1.9152 1.9571

a6 a7 a8 a9 a10

1.9785 1.9892 1.9946 1.9973 1.9986

(b) Let S be the set of positive integers for which an < 2. Then 1 ∈ S since

a1 =√

2 ∼= 1.4142 < 2.

Assume that k ∈ S. Since a2k+1 = 2ak < 4, it follows that ak+1 < 2. Thus k + 1 ∈ S and S is

the set of positive integers.

(c) 2 is the least upper bound.

(d) Let c be a positive number. Then c is the least upper bound of the set

S =

{√c,

√c√c,

√c

√c√c, . . .

}.

38. (a) a1∼= 1.4142136 a2

∼= 1.8477591 a3∼= 1.9615706 a4

∼= 1.9903695 a5∼= 1.9975909

a6∼= 1.9993976 a7

∼= 1.9998494 a8∼= 1.9999624 a9

∼= 1.9999906 a10∼= 1.9999976

(b) a1 =√

2 < 2. Assume true for an. Then an+1 =√

2 + an <√

2 + 2 = 2.

(c) lub S = 2.

(d) For any positive number c, lub S is the positive number satisfying

x =√c + x, that is, x = (1 +

√1 + 4c)/2

SECTION 11.2

1. an = 2 + 3(n− 1) = 3n− 1, n = 1, 2, 3, . . . 2. an = 1 − (−1)n, n = 1, 2, 3, . . .

3. an =(−1)n−1

2n− 1, n = 1, 2, 3, . . . 4. an =

2n − 12n

, n = 1, 2, 3, . . .

5. an =n2 + 1

n, n = 1, 2, 3, . . . 6. an = (−1)n

n

(n + 1)2, n = 1, 2, 3, . . .

7. an =

{n

1/n,

if n = 2k − 1

if n = 2k,where k = 1, 2, 3, . . .



SECTION 11.2 591

8. an =

{n

1/n2,

if n = 2k

if n = 2k − 1,where k = 1, 2, 3, . . .

9. decreasing; bounded below by 0 and above by 2

10. not monotonic; bounded below by −1 and above by 12 .

11.n + (−1)n

n= 1 + (−1)n

1n

: not monotonic; bounded below by 0 and above by32

12. increasing; bounded below by 1.001 but not bounded above.

13. decreasing; bounded below by 0 and above by 0.9

14. increasing; bounded below by 0 and above by 1

15.n2

n + 1= n− 1 +

1n + 1

: increasing; bounded below by12

but not bounded above

16. increasing; bounded below by√

2, but not bounded above:√n2 + 1 > n

17.4n√

4n2 + 1=

2√1 + 1/4n2

and1

4n2decreases to 0: increasing;

bounded below by 45

√5 and above by 2

18.2n

4n + 1=

2n

(2n)2 + 1=

12n + 1

2n

. decreasing; bounded below by 0 and above by 25 .

19. increasing; bounded below by 251 but not bounded above

20.n2

√n3 + 1

=1√

1n + 1

n4

and1n

+1n4

decreases to 0 =⇒ increasing; bounded below by√

22

,

but not bounded above.

21.2n

n + 1= 2 − 2

n + 1increases toward 2: increasing; bounded below by 0 and above by ln 2.

22. increasing (n ≥ 2); bounded below by2√

2310

, but not bounded above.


24. not monotonic; not bounded below and not bounded above

25. increasing; bounded below by√

3 and above by 2

26. decreasing (sincen + 1n

decreases to 1); bounded below by 0 and above by ln 2.



592 SECTION 11.2

27. (−1)2n+1√n = −√

n: decreasing; bounded above by −1 but not bounded below

28.√n + 1√n

=

√1 +

1n

decreasing; bounded below by 1 and above by√

2

29.2n − 1

2n= 1 − 1

2n: increasing; bounded below by

12

and above by 1

30.12n

− 12n + 3

=3

2n(2n + 3)decreasing; bounded below by 0 and above by 3

10 .

31. consider sinx as x → 0+: decreasing; bounded below by 0 and above by 1

32. not monotonic; bounded below by − 12 and above by 1

4


34. increasing (becausex

lnxis an increasing function on [4,∞).); bounded below by

4ln 4

but not bounded above.

35.1n− 1

n + 1=

1n (n + 1)

: decreasing; bounded below by 0 and above by12

36. not monotonic; bounded below by −1 and above by 1

37. Set f(x) =lnx

x. Then, f ′(x) =

1 − lnx

x2< 0 for x > e: decreasing;

bounded below by 0 and above by 13 ln 3.

38. not monotonic; not bounded below nor above (because exponentials grow faster than polynomials).

39. Set an =3n

(n + 1)2. Then,

an+1

an= 3

(n + 1n + 2

)2

> 1: increasing;

bounded below by 34 but not bounded above.

40.1 − ( 1

2 )n

( 12 )n

= 2n − 1 increasing; bounded below by 1 but not bounded above.

41. For n ≥ 5

an+1

an=

5n+1

(n + 1)!· n!5n

=5

n + 1< 1 and thus an+1 < an.

Sequence is not nonincreasing: a1 = 5 < 252 = a2.

42. For n ≥ M,an+1

an=

Mn+1

(n + 1)!· n!Mn

=M

n + 1< 1, so the sequence decreases for n ≥ M .



SECTION 11.2 593

43. boundedness: 0 < (cn + dn)1/n < (2dn)1/n = 21/nd ≤ 2d

monotonicity : an+1n+1 = cn+1 + d n+1 = ccn + ddn

< (cn + dn)1/ncn + (cn + dn)1/ndn

= (cn + dn)1+1/n

= (cn + dn)(n+1)/n

= an+1n

Taking the (n + 1)st root of each side we have an+1 < an. The sequence is decreasing.

44. If for all n

|an| ≤ M and |bn| ≤ N,

then for all n

|αan + βbn| ≤ |α||an| + |β||bn| ≤ |α|M + |β|N

and

|anbn| = |an||bn| ≤ MN

45. a1 = 1, a2 = 12 , a3 = 1

6 , a4 = 124 , a5 = 1

120 , a6 = 1720 ; an = 1

n!

46. a1 = 1, a2 = 8, a3 = 27, a4 = 64, a5 = 125, a6 = 216; an = n3

47. a1 = a2 = a3 = a4 = a5 = a6 = 1; an = 1

48. a1 = 1, a2 = 32 , a3 = 7

4 , a4 = 158 , a5 = 31

16 , a6 = 6332 ; an = (2n − 1)/2n−1

49. a1 = 1, a2 = 3, a3 = 5, a4 = 7, a5 = 9, a6 = 11; an = 2n− 1

50. a1 = 1, a2 = 12 , a3 = 1

3 , a4 = 14 , a5 = 1

5 , a6 = 16 ; an = 1/n

51. a1 = 1, a2 = 4, a3 = 9, a4 = 16, a5 = 25, a6 = 36; an = n2

52. a1 = 1, a2 = 3, a3 = 7, a4 = 15, a5 = 31, a6 = 63; an = 2n − 1

53. a1 = 1, a2 = 1, a3 = 2, a4 = 4, a5 = 8, a6 = 16; an = 2n−2 (n ≥ 3)

54. a1 = 3, a2 = 1, a3 = 3, a4 = 1, a5 = 3, a6 = 1; an = 2 − (−1)n

55. a1 = 1, a2 = 3, a3 = 5, a4 = 7, a5 = 9, a6 = 11; an = 2n− 1

56. a1 = 1, a2 = 3, a3 = 4, a4 = 5, a5 = 6, a6 = 7; an = n + 1 (n ≥ 2)

57. First a1 = 21 − 1 = 1. Next suppose ak = 2k − 1 for some k ≥ 1. Then

ak+1 = 2ak + 1 = 2(2k − 1

)+ 1 = 2k+1 − 1.



594 SECTION 11.2

58. True for n = 1. Assume true for n. Then an+1 = an + 5 = 5n− 2 + 5 = 5(n + 1) − 2.

59. First a1 =120

= 1. Next suppose ak =k

2k−1for some k ≥ 1. Then

ak+1 =k + 12k

ak =k + 12k

k

2k−1=

k + 12k

.

60. True for n = 1. Assume true for n. Then an+1 = an − 1n(n + 1)

=1n− 1

n(n + 1)=

n + 1 − 1n(n + 1)

=1

n + 1

61. (a) If r = 1 then Sn = n for n = 1, 2, 3, . . .

(b) Sn = 1 + r + r2 + · · · + rn−1

rSn = r + r2 + · · · + rn

Sn − rSn = 1 − rn

Sn =1 − rn

1 − r, r �= 1.

62.1

k(k + 1)=

1k− 1

k + 1, so

Sn = a1 + a2 + · · · + an−1 + an

=1

1 · 2 +1

2 · 3 + · · · + 1(n− 1)n

+1

n(n + 1)

=(

1 − 12

)+(

12− 1

3

)+ · · · +

(1

n− 1− 1

n

)+(

1n− 1

n + 1

)= 1 − 1

n + 1=

n

n + 1since all middle terms cancel out.

63. (a) Let Sn denote the distance traveled between the nth and (n + 1)st bounce. Then

S1 = 75 + 75 = 150, S2 = 34 (75) + 3

4 (75) = 150(

34

), . . . , Sn = 150

(34

)n−1

.

(b) An object dropped from rest from a height h feet above the ground will hit the ground in 14

√h

seconds. Therefore it follows that the ball will be in the air

Tn = 2(

14

)√Sn

2=

5√

32

(34

)(n−1)/2

seconds.

64. P0 = 5, P12 = 5 · 2, P24 = 5 · 22; so Pn = 5 2n/12

65. {an} is an increasing sequence; an → 12 as n → ∞.

66. {an} is a decreasing sequence; an → 2 as n → ∞.

67. (a) Let S be the set of positive integers for which an+1 > an. Since a2 = 1 +√a1 = 2 > 1, 1 ∈ S.

Assume that ak = 1 +√ak−1 > ak−1. Then

ak+1 = 1 +√ak > 1 +

√ak−1 = ak.

Thus, k ∈ S implies k + 1 ∈ S. It now follows that {an} is an increasing sequence.



SECTION 11.3 595

(b) Since {an} is an increasing sequence,

an = 1 +√an−1 < 1 +

√an, or an −√

an − 1 < 0.

Rewriting the second inequality as

(√an)2 −√

an − 1 < 0

and solving for√an it follows that

√an < 1

2 (1 +√

5). Hence, an < 12 (3 +

√5) for all n.

(c) a2 = 2, a3∼= 2.4142, a4

∼= 2.5538, a5∼= 2.5981, . . . , a9

∼= 2.6179, . . . , a15∼= 2.6180;

(e) lub {an} = 12 (3 +

√5) ∼= 2.6180

68. (a) We show that an < an+1 for all n. True for n = 1 since a1 = 1 <√

3 = a2. Assume true for

n, that is, an < an+1; we need to show that an+1 < an+2.

But an+1 =√

3an <√

3an+1 = an+2, as required.

(b) True since a1 < 3, and an < 3 =⇒ an+1 <√

3 · 3 = 3

(c) a1 = 1, a2 =√

3, a3∼= 2.2795, . . . , a14

∼= 2.9996, a15∼= 2.9998

(d) lub = 3

SECTION 11.3

1. diverges 2. converges to 0

3. converges to 0 4. diverges

5. converges to 1:n− 1n

= 1 − 1n→ 1 6. converges to 1:

n + (−1)n

n= 1 +

(−1)n

n→ 1

7. converges to 0:n + 1n2

=1n

+1n2

→ 0

8. converges to 0:π

2n→ 0, so sin

( π

2n

)→ sin 0 = 0

9. converges to 0: 0 <2n

4n + 1<

2n

4n=

12n

→ 0

10. diverges:n2

n + 1≥ n2

2n=

n

211. diverges

12. diverges:4n√n2 + 1

≈ 4n

n→ ∞

13. converges to 0

14. converges to 0:4n

2n + 106<

4n2n

=n

2n−2→ 0



596 SECTION 11.3

15. converges to 1:nπ

4n + 1→ π

4so tan

nπ

4n + 1→ tan

π

4= 1

16. converges to 0:1010

√n

n + 1=

1010

√n + 1/

√n→ 0

17. converges to49

:(2n + 1)2

(3n− 1)2=

4 + 4/n + 1/n2

9 − 6/n + 1/n2→ 4

9

18. converges to ln 2 :2n

n + 1→ 2, so ln

(2n

n + 1

)→ ln 2

19. converges to12

√2:

n2

√2n4 + 1

=1√

2 + 1/n4→ 1√

2

20. converges to 1:n4 − 1

n4 + n− 6=

1 − 1n4

1 + 1n3 − 6

n4

→ 1

21. diverges: cosnπ = (−1)n 22. diverges:n5

17n4 + 12= n

(1

17 + 12n4

)

23. converges to 1:1√n→ 0 so e1/

√n → e0 = 1

24. converges to√

4 = 2

25. converges to 0 : lnn− ln (n + 1) = ln(

n

n + 1

)→ ln 1 = 0

26. converges to 1:2n − 1

2n= 1 − 1

2n=⇒ 1

27. converges to12:

√n + 12√n

=12

√1 +

1n→ 1

228. converges to 0:

1n− 1

n + 1=

1n(n + 1)

→ 0

29. converges to e2:(

1 +1n

)2n

=[(

1 +1n

)n]2

→ e2

30. converges to√e :

(1 +

1n

)n/2

=

√(1 +

1n

)n

→√e

31. diverges; since 2n > n3 for n ≥ 10,2n

n2>

n3

n2= n

32. converges to ln 9 : 2 ln 3n− ln(n2 + 1) = ln(

9n2

n2 + 1

)→ ln 9

33. converges to 0:| sinn|√

n≤ 1√

n



SECTION 11.3 597

34. converges to π/4:n

n + 1→ 1, arctan 1 = π/4

35. converges to 1/2:√n2 + n− n =

(√n2 + n− n

) √n2 + n + n√n2 + n + n

=n√

n2 + n + n→ 1

2

36. converges to 2:√

4n2 + n

n=√

4 + (1/n) →√

4 = 2.

37. converges to π/2

38. converges to 31/9: let s = 3.444 · · ·. Then 10s− s = 31 and s = 31/9.

39. converges to −π/2:1 − n

n→ −1; arcsin (−1) = π/2.

40. converges to 1:(n + 1)(n + 4)(n + 2)(n + 3)

=n2 + 5n + 5n2 + 5n + 6

→ 1.

41. (a) n√n → 1 (b)

3n

n!→ 0

42. (a)n

n√n!

→ e (b) does not converge

43. b < n√an + bn = b n

√(a/b)n + 1 < b n

√2. Since 21/n → 1 as n → ∞, it follows that n

√an + bn → b

by the pinching theorem.

44. (a) −1 < r ≤ 1 (b) −1 < r < 1

45. Set ε > 0. Since an → L, there exists N1 such that

if n ≥ N1, then |an − L| < ε/2.

Since bn → M , there exists N2 such that

if n ≥ N2, then |bn −M | < ε/2.

Now set N = max {N1, N2}. Then, for n ≥ N ,

|(an + bn) − (L + M)| ≤ |an − L| + |bn −M | < ε

2+

ε

2= ε.

46. Let ε > 0, choose k such that n ≥ k =⇒ |an − L| < ε

|α| + 1.

Then for n ≥ k, |αan − αL| = |α||an − L| ≤ (|α| + 1)|an − L| < ε

Therefore αan → αL.

47. Since(

1 +1n

)→ 1 and

(1 +

1n

)n

→ e,

(1 +

1n

)n+1

=(

1 +1n

)n(1 +

1n

)→ (e)(1) = e.



598 SECTION 11.3

48. (a) If k = j, an =αk + αk−1 ·

1n

+ · · · + α0 ·1nk

βk + βk−1 ·1n

+ · · · + β0 ·1nk

→ αk

βk

(b) If k < j, an =αk + αk−1 ·

1n

+ · · · + α0 ·1nk

βj · nj−k + βj−1 · nj−1−k + · · · + β0 ·1nk

→ 0

(c) If k > j, an =αk · nk−j + αk−1 · nk−1−j + · · · + α0 ·

1nj

βj + βj−1 ·1n

+ · · · + β0 ·1nj

diverges

49. Suppose that {an} is bounded and non-increasing. If L is the greatest lower bound of the range of this

sequence, then an ≥ L for all n. Set ε > 0. By Theorem 11.1.4 there exists ak such that ak < L + ε.

Since the sequence is non-increasing, an ≤ ak for all n ≥ k. Thus,

L ≤ an < L + ε or |an − L| < ε for all n ≥ k

and an → L.

50. Let ε > 0. If an → L, then there exists a positive integer k such that

|an − L| < ε for all n ≥ k

If n ≥ k, then 2n ≥ k and 2n− 1 ≥ k, and thus

|en − L| = |a2n − L| < ε and |on − L| = |a2n−1 − L| < ε

It follows that en → L and on → L. If en → L and on → L, then there exist k1 and k2 such

that

if m ≥ k1, then |em − L| = |a2m − L| < ε

and

if m ≥ k2, then |om − L| = |a2m−1 − L| < ε

Let k = max{2k1, 2k2 − 1}. If n ≥ k then

either an = a2m with m > k1 or an = a2m−1 with m ≥ k2

In either case, |an − L| < ε. This shows that an → L.

51. Let ε > 0. Choose k so that, for n ≥ k,

L− ε < an < L + ε, L− ε < cn < L + ε and an ≤ bn ≤ cn.

For such n,

L− ε < bn < L + ε.

52. Let M be a bound for {bn}. Then |anbn| ≤ |an|M.

Given ε > 0, choose k such that |an| < ε/M for n ≥ k. Then |anbn| < ε for n ≥ k.



SECTION 11.3 599

53. Let ε > 0. Since an → L, there exists a positive integer N such that L− ε < an < L + ε for all n ≥ N.

Now an ≤ M for all n, so L− ε < M, or L < M + ε. Since ε is arbitrary, L ≤ M.

54. The converse is false. For example, let an = (−1)n. Then |an| → 1, but {an} diverges.

55. Assume an → 0 as n → ∞. Let ε > 0. There exists a positive integer N such that |an − 0| < ε for

all n ≥ N . Since | |an| − 0| ≤ |an − 0|, it follows that |an| → 0. Now assume that |an| → 0. Since

−|an| ≤ an ≤ |an|, an → 0 by the pinching theorem.

56. Let ε > 0. There exists a positive integer N1 such that |an − L| < ε for all n > 2N1 − 1, and there

exists a positive integer N2 such that |bn − L| < ε for all n > 2N2. The sequence a1, b1, a2, b2, . . .

can be represented by the sequence c1, c2, c3, . . ., where

cn =

{a(n+1)/2, if n is odd

bn/2, if n is even.

}

Let N = max{2N1 − 1, 2N2}. Then n > N =⇒ |cn − L| < ε =⇒ cn → L.

57. By the continuity of f, f(L) = f(

limn→∞

an

)= lim

n→∞f(an) = lim

n→∞an+1 = L.

58.2n

n!=

21· 22· 23· · · 2

n= 2 · 2

n· (terms that are ≤ 1) ≤ 4

n.

Since4n→ 0 and 0 <

2n

n!≤ 4

n,

2n

n!→ 0 as well.

59. Set f(x) = x1/p. Since1n→ 0 and f is continuous at 0, it follows by Theorem 11.3.12 that(

1n

)1/p

→ 0.

60. Since |an − L| = |(an − L) − 0| = ||an − L| − 0|,|an − L| < ε iff |(an − L) − 0| < ε iff ||an − L| − 0| < ε,

So an → L iff an − L → 0 iff |an − L| → 0.

61. an = e1−n → 0 62. diverges

63. an =1n!

→ 0 64. an = 1 · 12· 23· · · n− 1

n=

1n

converges to 0

65. an =12[1 − (−1)n] diverges 66. an =

2n − 12n−1

→ 2

67. L = 0, n = 32 68.1√n→ 0.

1√n< 0.001 for n ≥ 10002 + 1

69. L = 0, n = 4 70.n10

10n→ 0.

n10

10n< 0.001 for n ≥ 15



600 SECTION 11.3

71. L = 0, n = 7 72.2n

n!→ 0.

2n

n!< 0.001 for n ≥ 10

73. L = 0, n = 65 74.lnn

n→ 0.

lnn

n< 0.001 for n ≥ 9119

75. (a) an+1 = 1 +√an Suppose that an → L as n → ∞. Then an+1 → L as n → ∞. Therefore

L = 1 +√L which, since L > 1, implies L = 1

2 (3 +√

5).

(b) an+1 =√

3an Suppose that an → L as n → ∞. Then an+1 → L as n → ∞. Therefore

L =√

3L which, since L > 1, implies L = 3.

76. (a) a2∼= 2.6458, a3

∼= 2.9404, a4∼= 2.9900, a5

∼= 2.9983, a6∼= 2.9997

(b) True for n = 1. Assume true for n. Then an+1 =√

6 + an−1 ≤√

6 + 3 = 3

(c) an+12 − an

2 = 6 + an − an2 = (3 − an)(2 + an) ≥ 0 since 0 ≤ an ≤ 3.

Since ak ≥ 0 for all k, this implies an+1 ≥ an

(d) an → 3

77. (a) a2 a3 a4 a5 a6 a7 a8 a9 a10

0.5403 0.8576 0.6543 0.7935 0.7014 0.7640 0.7221 0.7504 0.7314

(b) L is a fixed point of f(x) = cosx, that is, cosL = L; L ∼= 0.739085.

78. (a) a2∼= 1.5403, a3

∼= 1.5708, a4∼= a5

∼= · · · ∼= a10∼= 1.5708.

(b) L ∼= 1.570796 Let f(x) = x + cosx. L must satisfy L = f(L), so L = L + cosL,

and cosL = 0. Indeed, the L we found is justπ

2∼= 1.570796327

PROJECT 11.3

1. (a)a2 a3 a4 a5 a6 a7 a8

2.000000 1.750000 1.732143 1.732051 1.732051 1.732051 1.732051

(b) L =12

(L +

3L

)which implies L2 = 3 or L =

√3.

2. The Newton-Raphson method applied to the function f(x) = x2 −R gives

an = an−1 −f(an−1)f ′(an−1)

= an−1 −a2n−1 −R

2an−1

=12an−1 +

12

R

an−1=

12

(an−1 +

R

an−1

), n = 2, 3, . . . .

3 & 4. (a) f(x) = x3 − 8, so xn → 2 (b) f(x) = sinx− 12 , so xn → π

6

(c) f(x) = lnx− 1, so xn → e



SECTION 11.4 601

SECTION 11.4

1. converges to 1: 22/n = (21/n)2 → 12 = 1 2. converges to 1: e−α/n → e0 = 1

3. converges to 0: for n > 3, 0 <

(2n

)n

<

(23

)n

→ 0

4. converges to 0:log10 n

n=

1ln 10

· lnn

n→ 0

5. converges to 0:ln (n + 1)

n=[ln (n + 1)n + 1

](n + 1n

)→ (0)(1) = 0

6. converges to 0:3n

4n=(

34

)n

→ 0 7. converges to 0:x100n

n!=

(x100)n

n!→ 0

8. converges to 1: n1/(n+2) =(n1/n

)n/(n+2)

→ 1 9. converges to 1: nα/n = (n1/n)α → 1α = 1

10. converges to 0: ln(n + 1n

)→ ln(1) = 0

11. converges to 0:3n+1

4n−1= 12

(3n

4n

)= 12

(34

)n

→ 12(0) = 0

12. converges to12

:∫ 0

−n

e2x dx =12− e−2n

2→ 1

2

13. converges to 1: (n + 2)1/n = e1n ln(n+2) and, since

1n

ln (n + 2) =[ln (n + 2)n + 2

](n + 2n

)→ (0)(1) = 0,

it follows that (n + 2)1/n → e0 = 1.

14. converges to e−1 :(

1 − 1n

)n

=(

1 +(−1)n

)n

→ e−1 (by (11.4.7)

15. converges to 1:∫ n

0

e−x dx = 1 − 1en

→ 1 16. diverges.

17. converges to π: integral = 2∫ n

0

dx

1 + x2= 2 tan−1 n → 2

(π2

)= π

18. converges to 0:∫ n

0

e−nx dx = −e−n2

n+

1n→ 0

19. converges to 1: recall (11.4.6) 20. converges to 0: n2 sinnπ = 0 for all n

21. converges to 0:ln (n2)

n= 2

ln n

n→ 2(0) = 0



602 SECTION 11.4

22. converges to π :∫ 1−1/n

−1+1/n

dx√1 − x2

= sin−1

(1 − 1

n

)− sin−1

(−1 +

1n

)→ sin−1(1) − sin−1(−1) = π

23. diverges: Since limx→0

sinx

x= 1,

n

πsin

π

n=

sin (π/n)π/n

→ 1. Therefore,

n2 sinπ

n= nπ

(nπ

sinπ

n

)→ nπ.

24. diverges 25. converges to 0:5n+1

42n−1= 20

(516

)n

→ 0

26. converges to e3x :(1 +

x

n

)3n

=[(

1 +x

n

)n]3→ (ex)3 = e3x

27. converges to e−1:(n + 1n + 2

)n

=(

1 − 1n + 2

)n

=

(1 +

(−1)n + 2

)n+2

(1 +

(−1)n + 2

)2 → e−1

1= e−1

28. converges to 2:∫ 1

1/n

dx√x

= 2 − 2√n→ 2.

29. converges to 0: 0 <

∫ n+1

n

e−x2dx ≤ e−n2

[(n + 1) − n] = e−n2 → 0

30. converges to 1:(

1 +1n2

)n

=

[(1 +

1n2

)n2]1/n

→ (e1)0 = 1

31. converges to 0:nn

2n2 =( n

2n)n

→ 0 sincen

2n→ 0

32. converges to 0:∫ 1/n

0

cos ex dx →∫ 0

0

cos ex dx = 0

33. converges to ex: use (11.4.7)

34. diverges:(

1 +1n

)n2

=[(

1 +1n

)n]n> 2n,

[(1 +

1n

)n

≈ e > 2]

35. converges to 0:

∣∣∣∣∣∫ 1/n

−1/n

sinx2dx

∣∣∣∣∣ ≤∫ 1/n

−1/n

| sin x2| dx ≤∫ 1/n

−1/n

1 dx =2n→ 0

36.(t +

x

n

)n= tn

(1 +

x/t

n

)n

; converges to 0 if t < 1, converges to ex if t = 1, diverges if t > 1.

37. converges:sin(6/n)sin(3/n)

→ 2 38. converges:arctann

n→ 0



SECTION 11.4 603

39.√n + 1 −

√n =

√n + 1 −√

n√n + 1 +

√n

(√n + 1 +

√n)

=1√

n + 1 +√n→ 0

40.√n2 + n− n =

√n2 + n− n√n2 + n + n

(√n2 + n + n) =

n√n2 + n + n

=1

1 +√

1 + 1/n→ 1

2

41. (a) The length of each side of the polygon is 2r sin(π/n). Therefore the perimeter, pn, of the polygon

is given by: pn = 2rn sin(π/n).

(b) 2rn sin(π/n) → 2πr as n → ∞ : The number 2rn sin (π/n) is the perimeter of a regular polygon

of n sides inscribed in a circle of radius r. As n tends to ∞, the perimeter of the polygon tends

to the circumference of the circle.

42. Since 0 < c < d, d < (cn + dn)1/n < (2dn)1/n = 21/nd → d, so by the pinching theorem

(cn + dn)1/n → d.

43. By the hint, limn→∞

1 + 2 + . . . + n

n2= lim

n→∞n(n + 1)

2n2= lim

n→∞1 + 1/n

2=

12.

44. diverges:12 + 22 + · · · + n2

(1 + n)(2 + n)=

n(n + 1)(2n + 1)6(1 + n)(2 + n)

=2n3 + 3n2 + n

6n2 + 18n + 12→ ∞

45. By the hint, limn→∞

13 + 23 + . . . + n3

2n4 + n− 1= lim

n→∞n2(n + 1)2

4(2n4 + n− 1)= lim

n→∞1 + 2/n + 1/n2

8 + 4/n3 − 4/n4=

18.

46. Here we show that every convergent sequence is a Cauchy sequence. Let ε > 0. If an → L, then there

exists a positive integer k such that

|ap − L| < ε

2for all p ≥ k

With m,n ≥ k we have

|am − an| ≤ |am − L| + |L− an| = |am − L| + |an − L| < ε

2+

ε

2= ε.

47. (a) mn+1 −mn =1

n + 1(a1 + · · · + an + an+1) −

1n

(a1 + · · · + an)

=1

n(n + 1)[nan+1 −

n

(

︷︸︸︷a1 + · · · + an)

]> 0 since {an} is increasing.

(b) We begin with the hint

mn <|a1 + · · · + aj |

n+

ε

2

(n− j

n

).

Since j is fixed,

|a1 + · · · + aj |n

→ 0



604 SECTION 11.4

and therefore for n sufficiently large

|a1 + · · · + aj |n

<ε

2.

Sinceε

2

(n− j

n

)<

ε

2,

we see that, for n sufficiently large, |mn| < ε. This shows that mn → 0.

48. (a) Since {an} converges, it is a Cauchy sequence (see Exercise 46), so given ε > 0 we can find k

such that |an − am| < ε for n,m ≥ k.

In particular, |an+1 − an| < ε, so limn→∞

(an − an−1) = 0.

(b) {an} does not necessarily converge. For example let an = lnn. This diverges, but

an − an−1 = lnn− ln(n− 1) = ln(

n

n− 1

)→ ln 1 = 0

49. (a) Let S be the set of positive integers n (n ≥ 2) for which the inequalities hold. Since(√b)2

− 2√ab +

(√a)2 =

(√b−

√a)2

> 0,

it follows thata + b

2>

√ab and so a1 > b1. Now,

a2 =a1 + b1

2< a1 and b2 =

√a1b1 > b1.

Also, by the argument above,

a2 =a1 + b1

2>√

a1b1 = b2,

and so a1 > a2 > b2 > b1. Thus 2 ∈ S. Assume that k ∈ S. Then

ak+1 =ak + bk

2<

ak + ak2

= ak, bk+1 =√

akbk >√b2k = bk,

and

ak+1 =ak + bk

2>√

akbk = bk+1.

Thus k + 1 ∈ S. Therefore, the inequalities hold for all n ≥ 2.

(b) {an} is a decreasing sequence which is bounded below.

{bn} is an increasing sequence which is bounded above.

Let La = limn→∞

an, Lb = limn→∞

bn. Then

an =an−1 + bn−1

2implies La =

La + Lb

2and La = Lb.

50.e−

(1 + 1

100

)100e

∼= 0.004995 : within 0.01%;e5 −

(1 + 5

100

)100e5

∼= 0.11395 : within 12%

e−(1 + 1

1000

)1000e

∼= 0.0004995 : within 0.05%;e5 −

(1 + 5

1000

)1000e5

∼= 0.01238 : within 1.3%



SECTION 11.5 605

51. The numerical work suggests L ∼= 1. Justification: Set f(x) = sinx− x2. Note that f(0) = 0 and

f ′(x) = cosx− 2x > 0 for x close to 0. Therefore sinx− x2 > 0 for x close to 0 and sin 1/n− 1/n2 > 0

for n large. Thus, for n large,

1n2

< sin1n<

1n

(| sinx| ≤ |x| for all x)

(1n2

)1/n

<

(sin

1n

)1/n

<

(1n

)1/n

(1

n1/n

)2

<

(sin

1n

)1/n

<1

n1/n.

As n → ∞ both bounds tend to 1 and therefore the middle term also tends to 1.

52. Numerical work suggests L ∼= 1/3. Conjecture: L = 1/k.

Proof: (nk + nk−1)1/k − n = n(1 + 1/n)1/k − n =(1 + 1/n)1/k − 1

1/n;

limn→∞

(1 + 1/n)1/k − 11/n

= limh→0

(1 + h)1/k − 1h

= f ′(1),

where f(x) = x1/k. Since f ′(x) = (1/k)x(1/k)−1, f ′(1) = 1/k.

53. (a) a3 a4 a5 a6 a7 a8 a9 a10

2 3 5 8 13 21 34 55

(b) r1 r2 r3 r4 r5 r6

1 2 1.5 1.667 1.600 1.625

(c) Following the hint,

1 +1

rn−1= 1 +

1anan−1

= 1 +an−1

an=

an + an−1

an=

an+1

an= rn.

Now, if rn → L, then rn−1 → L and

1 +1L

= L which, since L > 1, implies L =1 +

√5

2∼= 1.618034.

54. With the partition {0, 1n ,

2n , . . . ,

nn} and f(x) = x, we have

an =1n

(1n

+2n

+ · · · + n

n

)=

1n

[f

(1n

)+ f

(2n

)+ · · · + f

(nn

)]=

n∑i=1

f(xi)Δxi,

so it is a Riemann sum for∫ 1

0

x dx, and therefore limn→∞

an =∫ 1

0

x dx =12

SECTION 11.5

(We’ll use � to indicate differentiation of numerator and denominator.)

1. limx→0+

sin x√x

�= limx→0+

2√x cos x = 0 2. lim

x→1

lnx

1 − x

�= limx→1

1/x−1

= −1



606 SECTION 11.5

3. limx→0

ex − 1ln (1 + x)

�= limx→0

(1 + x)ex = 1 4. limx→4

√x− 2x− 4

�= limx→4

12√x

1=

14

5. limx→π/2

cos x

sin 2x�= lim

x→π/2

− sin x

2 cos 2x=

12

6. limx→a

x− a

xn − an�= lim

x→a

1nxn−1

=1

nan−1

7. limx→0

2x − 1x

�= limx→0

2x ln 2 = ln 2 8. limx→0

tan−1 x

x

�= limx→0

11+x2

1= 1

9. limx→1

x1/2 − x1/4

x− 1�= lim

x→1

(12x−1/2 − 1

4x−3/4

)=

14

10. limx→0

ex − 1x(1 + x)

�= limx→0

ex

1 + 2x= 1

11. limx→0

ex − e−x

sin x

�= limx→0

ex + e−x

cos x= 2 12. lim

x→0

1 − cosx3x

�= limx→0

sinx

3= 0

13. limx→0

x + sinπx

x− sinπx

�= limx→0

1 + π cosπx1 − π cosπx

=1 + π

1 − π

14. limx→0

ax − (a + 1)x

x

�= limx→0

ax ln a− (a + 1)x ln(a + 1)1

= ln(

a

a + 1

)

15. limx→0

ex + e−x − 21 − cos 2x

�= limx→0

ex − e−x

2 sin 2x�= lim

x→0

ex + e−x

4 cos 2x=

12

16. limx→0

x− ln(x + 1)1 − cos 2x

�= limx→0

1 − 1x+1

2 sin 2x�= lim

x→0

1(x+1)2

4 cos 2x=

14

17. limx→0

tanπx

ex − 1�= lim

x→0

π sec2 πx

ex= π

18. limx→0

cosx− 1 + x2/2x4

�= limx→0

− sinx + x

4x3

�= limx→0

− cosx + 112x2

�= limx→0

sinx

24x=

124

19. limx→0

1 + x− ex

x(ex − 1)�= lim

x→0

1 − ex

xex + ex − 1�= lim

x→0

−ex

xex + 2ex= −1

2

20. limx→0

ln(secx)x2

�= limx→0

tanx

2x�= lim

x→0

sec2 x

2=

12

21. limx→0

x− tan x

x− sin x

�= limx→0

1 − sec2 x

1 − cos x

�= limx→0

−2 sec2 x tan x

sin x= lim

x→0

−2 sec2 x

cos x= −2

22. limx→0

xenx − x

1 − cosnx�= lim

x→0

enx + nxenx − 1n sinnx

�= limx→0

enx(2n + n2x)n2 cosnx

=2n

23. limx→1−

√1 − x2

√1 − x3

= limx→1−

√1 − x2

1 − x3=

√23

=13

√6 since lim

x→1−

1 − x2

1 − x3

�= limx→1−

2x3x2

=23



SECTION 11.5 607

24. limx→0

2x− sinπx

4x2 − 1= 0

25. limx→π/2

ln (sin x)(π − 2x)2

�= limx→π/2

− cot x

4(π − 2x)�= lim

x→π/2

csc2 x

−8= −1

8

26. limx→0+

√x√

x + sin√x

�= limx→0+

12√x

12√x

+ cos√x

2√x

= limx→0+

11 + cos

√x

=12

27. limx→0

cosx− cos 3xsin(x2)

�= limx→0

− sinx + 3 sin 3x2x cos(x2)

�= limx→0

− cosx + 9 cos 3x2 cos(x2) − 4x2 sin(x2)

= 4

28. limx→0

√a + x−√

a− x

x

�= limx→0

12√a+x

+ 12√a−x

1=

1√a

=√a

a

29. limx→π/4

sec2 x− 2 tan x

1 + cos 4x�= lim

x→π/4

2 sec2 x tan x− 2 sec2 x

−4 sin 4x

�= limx→π/4

2 sec4 x + 4 sec2 x tan2 x− 4 sec2 x tan x

−16 cos 4x=

12

30. limx→0

x− arcsinx

sin3 x

�= limx→0

1 − 1√1−x2

3 sin2 x cosx�= lim

x→0

−x(1−x2)3/2

6 sinx cos2 x− 3 sin3 x

�= limx→0

−1−2x2

(1−x2)5/2

6 cos3 x− 21 sin2 x cosx= −1

6

31. limx→0

tan−1 x

tan−1 2x�= lim

x→0

11 + x2

˙1 + 4x2

2=

12

32. limx→0

sin−1 x

x

�= limx→0

1√1−x2

1= 1

33. 1 : limx→∞

π/2 − tan−1 x

1/x�= lim

x→∞x2

1 + x2= 1

34. −1 : limn→∞

ln(1 − 1n )

sin( 1n )

= limx→0+

ln(1 − x)sinx

�= limx→0+

1(1 − x) cosx

= −1

35. 1 : limx→∞

1x[ ln (x + 1) − ln x ]

= limx→∞

1/xln (1 + 1/x)

= limt→0+

t

ln (1 + t)�= lim

t→0+(1 + t) = 1

36.13

: limn→∞

sinh(π/n) − sin(π/n)sin3(π/n)

= limx→∞

sinh(π/x) − sin(π/x)sin3(π/x)

�= limu→0+

sinhu− sinu

sin3 u

�= limu→0+

coshu− cosu3 sin2 u cosu

�= limu→0+

sinhu + sinu

6 sinu cos2 u− 3 sin3 u

�= limu→0+

coshu + cosu6 cos3 u− 21 sin2 u cosu

=26

=13



608 SECTION 11.5

37. limx→0

x3

4x − 1= 0 38. lim

x→0

4xsin2 x

does not exist

39. limx→0

xπ2 − arccosx

= 1 40. limx→2+

√2x− 2√x− 2

= 0

41. limx→0

tanhx

x= 1 42. lim

x→π/2

1 + cos 2x1 − sinx

= 4

43. limx→0

(2 + x + sin x) �= 0, limx→0

(x3 + x− cos x) �= 0

44. limn→∞

n(a1/n − 1

)= lim

x→∞a1/x − 1

1/x�= lim

x→∞

a1/x ln a(− 1

x2

)(− 1

x2

) = ln a

45. The limit does not exist if b �= 1. Therefore, b = 1.

limx→0

cos ax− 12x2

�= limx→0

−a sin ax

4x�= lim

x→0

−a2 cos ax4

= − a2

4

Now, − a2

4= − 4 implies a = ±4.

46. limx→0

sin 2x + ax + bx3

x3

�= limx→0

2 cos 2x + a + 3bx2

3x2need a = −2 to keep numerator 0

�= limx→0

−4 sin 2x + 6bx6x

�= limx→0

−8 cos 2x + 6b6

= 0 if 6b = 8

=⇒ a = −2, b =43

47. Recall limx→0+

(1 + x)1/x = e:

limx→0+

(1 + x)1/x − e

x

�= limx→0+

[(1 + x)1/x

] [x− (1 + x) ln (1 + x)x2 + x3

]

= e limx→0+

x− (1 + x) ln (1 + x)x2 + x3

�= e limx→0+

− ln (1 + x)2x + 3x2

�= e limx→0+

−1/(1 + x)2 + 6x

= −e

2

48. (a) limh→0

f(x + h) − f(x− h)2h

�= limh→0

f ′(x + h) − f ′(x− h)(−1)2

= f ′(x)

(note that here we differentiated with respect to h, not x. )

(b) limh→0

f(x + h) − 2f(x) + f(x− h)h2

�= limh→0

f ′(x + h) − f ′(x− h)2h

�= limh→0

f ′′(x + h) + f ′′(x− h)2

= f ′′(x)



SECTION 11.5 609

49. limx→0

1x

∫ x

0

f(t) dt �= limx→0

f(x)1

= f(0)

50. (a) limx→0

Si(x)x

�= limx→0

sinx

x= 1

(b) limx→0

Si(x) − x

x3

�= limx→0

sinx/x− 13x2

= limx→0

sinx− x

3x3

�= limx→0

cosx− 19x2

�= limx→0

− sinx

18x= − 1

18

51. (a) limx→0

C(x)x

�= limx→0

cos2 x1

= 1

(b) limx→0

C(x) − x

x3

�= limx→0

cos2 x− 13x2

�= limx→0

−2 cosx sinx

6x�= lim

x→0

−2 cos2 x + 2 sin2 x

6= −1

3

52. (a) limx→a

∫ x

a

f(t) dt

f(x)�= lim

x→a

f(x)f ′(x)

= 0

(b) Similarly, limx→a

∫ x

a

f(t) dt

f(x)�= lim

x→a

f(x)f ′(x)

�= · · · �= limx→a

f (k−1)(x)f (k)(x)

= 0

53. A(b) = 2∫ √

b

0

(b− x2) dx = 2[bx− x3

3

]√b

0

=43b√b and T (b) =

12

(2√b)b = b

√b.

Thus, limb→0

T (b)A(b)

=b√b

43 b

√b

=34.

54. T (θ) =12(1 − cos θ) sin θ; S(θ) =

θ

2− 1

2sin θ:

limθ→0+

T (θ)S(θ)

= limθ→0+

(1 − cos θ) sin θ

θ − sin θ

�= limθ→0+

(1 − cos θ) cos θ + sin2 θ

1 − cos θ= lim

θ→0+

cos θ − cos 2θ1 − cos θ

�=− sin θ + 2 sin 2θ

sin θ= lim

θ→0+

− sin θ + 4 sin θ cos θsin θ

= 3

55. (a) f(x) → ∞ as x → ±∞

-5 5x

5

10

15

y

f

(b) f(x) → 10 as x → 4

Confirmation: limx→4

x2 − 16√x2 + 9 − 5

�= limx→4

2x

x (x2 + 9)−1/2= lim

x→42√x2 + 9 = 10



610 SECTION 11.6

56. (a) limx→±∞

f(x) = 0

−10 10x

0.17

y

(b) f(x) → 16

as x → 0; limx→0

x− sinx

x3

�= limx→0

1 − cosx3x2

�= limx→0

sinx

6x=

16

57. (a) f(x) → 0.7 as x → 0

-2 -1 1 2x

0.6

0.8

y

f

(b) Confirmation: limx→0

2sinx − 1x

�= limx→0

ln(2) 2sinx cosx1

= ln 2 ∼= 0.6931

58. (a) g(x) → −1.6 as x → 0x

− 1

− 2

y

(b) Confirmation: limx→0

3cosx − 3x2

�= limx→0

3cosx(− sinx) ln 32x

= − limx→0

3cosx sinx

2xln 3

= − 3 ln 32

∼= −1.6479

SECTION 11.6

(We’ll use � to indicate differentiation of numerator and denominator.)

1. limx→−∞

x2 + 11 − x

�= limx→−∞

2x−1

= ∞ 2. limx→∞

20xx2 + 1

= 0

3. limx→∞

x3

1 − x3= lim

x→∞1

1/x3 − 1= −1 4. lim

x→∞x3 − 12 − x

= −∞

5. limx→∞

x2 sin1x

= limh→0+

[(1h

)(sinh

h

)]= ∞



SECTION 11.6 611

6. limx→∞

ln(xk)x

�= limx→∞

k/x

1= 0

7. limx→π

2−

tan 5xtanx

= limx→π

2−

[(sin 5xsinx

)( cosxcos 5x

)]=

15

since

limx→π

2−

sin 5xsinx

= 1 and limx→π

2−

cosxcos 5x

�= limx→π

2−

sinx

5 sin 5x=

15

8. limx→0

(x ln | sinx|) = limx→0

ln | sinx|1/x

�= limx→0

cosxsinx

− 1x2

= limx→0

(− x

sinx

)(x cosx) = 0

9. limx→0+

x2x = limx→0+

(xx)2 = 12 = 1 [see (11.6.4)]

10. limx→∞

x sinπ

x= lim

t→0+

sinπt

t

�= limt→0+

π cosπt1

= π

11. limx→0

x( ln |x| )2 = limx→0

(ln |x|)21/x

�= limx→0

2 ln |x|−1/x

�= limx→0

21/x

= limx→0

2x = 0

12. limx→0+

lnx

cotx�= lim

x→0+

1/x− csc2 x

= limx→0+

− sin2 x

x= lim

x→0+− sinx

x· sinx = 0

13. limx→∞

1x

∫ x

0

et2dt

�= limx→∞

ex2

1= ∞

14. limx→∞

√1 + x2

x= lim

x→∞

√1x2

+ 1 = 1

15. limx→0

[1

sin2 x− 1

x2

]= lim

x→0

x2 − sin2 x

x2 sin2 x

�= limx→0

2x− 2 sinx cosx2x2 sinx cosx + 2x sin2 x

= limx→0

2x− sin 2xx2 sin 2x + 2x sin2 x

�= limx→0

2 − 2 cos 2x2x2 cos 2x + 4x sin 2x + 2 sin2 x

�= limx→0

4 sin 2x−4x2 sin 2x + 12x cos 2x + 6 sin 2x

�= limx→0

8 cos 2x−8x2 cos 2x− 32x sin 2x + 24 cos 2x

=13

16. Since limx→0

ln(| sinx|x) = limx→0

(x ln | sinx|) = 0 by Exercise 8, limx→0

| sinx|x = e0 = 1

17. limx→1

x1/(x−1) = e since limx→1

ln[x1/(x−1)

]= lim

x→1

ln x

x− 1�= lim

x→1

1x

= 1

18. Take log:

limx→0+

ln(xsinx

)= lim

x→0+(sinx lnx) = lim

x→0+

(lnx

cscx

)�= lim

x→0+

1/x− cscx cotx

= limx→0+

− sin2 x

x cosx= 0, so lim

x→0+xsinx = e0 = 1



612 SECTION 11.6

19. limx→∞

(cos

1x

)x

= 1 since limx→∞

ln[(

cos1x

)x]= lim

x→∞

ln(

cos1x

)(1/x)

�= limx→∞

(− sin (1/x)

cos (1/x)

)= 0

20. Take log:

limx→π/2

ln (| secx|cosx) = limx→π/2

cosx ln | secx| = limx→π/2

ln | secx|secx

�= limx→π/2

tanx

secx tanx= lim

x→π/2cosx = 0, so lim

x→π/2| secx|cosx = e0 = 1

21. limx→0

[1

ln (1 + x)− 1

x

]= lim

x→0

x− ln (1 + x)x ln (1 + x)

�= limx→0

x

x + (1 + x) ln (1 + x)

�= limx→0

11 + 1 + ln (1 + x)

=12

22. Take log: limx→∞

ln(x2 + a2)(1/x)2 = limx→∞

ln(x2 + a2)x2

�= limx→∞

2xx2+a2

2x= 0,

so limx→∞

(x2 + a2)(1/x)2 = e0 = 1

23. limx→0

[1x− cotx

]= lim

x→0

sinx− x cosxx sinx

�= limx→0

x sinx

sinx + x cosx

�= limx→0

sinx + x cosx2 cosx− x sinx

= 0

24. limx→∞

ln(x2 − 1x2 + 1

)3

= 3 limx→∞

ln(x2 − 1x2 + 1

)= 0

25. limx→∞

(√x2 + 2x− x

)= lim

x→∞

[(√x2 + 2x− x

)(√x2 + 2x + x√x2 + 2x + x

)]

= limx→∞

2x√x2 + 2x + x

= limx→∞

2√1 + 2/x + 1

= 1

26. limx→∞

(1 +

a

x

)bx= lim

x→∞

[(1 +

a

x

)x]b= (ea)b = eab.

27. limx→∞

(x3 + 1

)1/ lnx = e3 since

limx→∞

ln[(x3 + 1

)1/ lnx]

= limx→∞

ln(x3 + 1

)lnx

�= limx→∞

(3x2

x3 + 1

)1/x

= limx→∞

31 + 1/x3

= 3.


ln(ex + 1)x

�= limx→∞

ex

ex+1

1�= lim

x→∞ex

ex= 1,

so limx→∞

(ex + 1)1/x = e



SECTION 11.6 613

29. limx→∞

(coshx)1/x = e since

limx→∞

ln [(coshx)1/x] = limx→∞

ln (coshx)x

�= limx→∞

sinhx

coshx= 1.


3x ln(

1 +1x

)= 3 lim

x→∞

ln(1 + 1

x

)1x

�= 3 limx→∞

−1/x2

1+1/x

− 1x2

= 3,

so limx→∞

(1 +

1x

)3x

= e3

31. limx→0

(1

sinx− 1

x

)= lim

x→0

x− sinx

x sinx

�= limx→0

1 − cosxsinx + x cosx

�= limx→0

sinx

2 cosx− x sinx= 0

32. Take log: limx→0

ln(ex + 3x)x

�= limx→0

ex+3ex+3x

1= 4, so lim

x→0(ex + 3x)1/x = e4

33. limx→1

(1

lnx− x

x− 1

)= lim

x→1

x− 1 − x lnx

(x− 1) lnx

�= limx→1

− lnx

(x− 1)(1/x) + lnx

= limx→1

−x lnx

x− 1 + x lnx

�= limx→1

− lnx− 12 + lnx

= − 12

34.√

2: take log: limx→0

ln(1 + 2x) − ln 2x

�= limx→0

2x ln 21 + 2x

1=

ln 22

; limx→0

(1 + 2x

2

)1/x

= e12 ln 2 =

√2

35. 0 :1n

ln1n

= − lnn

n→ 0 36. 0: lim

n→∞nk

2n→ 0

37. 1 : ln[(lnn)1/n

]=

1n

ln (lnn) → 0

38. 0: limn→∞

lnn

np

�= limn→∞

1/npnp−1

= limn→∞

1p np

= 0

39. 1 : ln[(n2 + n

)1/n] =1n

ln [n(n + 1)] =lnn

n+

ln (n + 1)n

→ 0

40. 1: limn→∞

ln(nsin(π/n)

)= lim

n→∞[sin(π/n) lnn] = lim

n→∞

(sin(π/n)

1/n

) (lnn

n

)= 0, so nsin(π/n) → 1

41. 0 : 0 ≤ n2 lnn

en<

n3

en, lim

x→∞x3

ex= 0

42. 1: take log: ln(√n− 1)1/

√n =

ln(√n− 1)√n

→ 0, so (√n− 1)1/

√n → 1

43. limx→0

(sinx)x = 1 44. limx→π/4

(tanx)tan2x =1e



614 SECTION 11.6

45. limx→0

(1

sinx− 1

tanx

)= 0 46. lim

x→0+(sinhx)−x = 1

47.

vertical asymptote y-axis

48.

vertical asymptote x = 1

horizontal asymptote y = 1

49.

horizontal asymptote x-axis

50.


51.


52.

vertical asymptote y-axis


53.b

a

√x2 − a2 − b

ax =

√x2 − a2 + x√x2 − a2 + x

(b

a

)(√x2 − a2 − x

)=

−ab√x2 − a2 + x

→ 0 as x → ∞

54. coshx− sinhx =12(ex + e−x) − 1

2(ex − e−x) = e−x → 0, as x → ∞

55. for instance, f(x) = x2 +(x− 1)(x− 2)

x3

56. for instance, F (x) = x +sinx

1 + x2

57. limx→0+

− 2xcos x

�= limx→0+

2− sin x

. L’Hospital’s rule does not apply here since limx→0+

cos x = 1.



SECTION 11.6 615

58. (a) Let S be the set of positive integers for which the statement is true. Since limx→∞

lnx

x= 0, 1 ∈ S.

Assume that k ∈ S. By L’Hospital’s rule,

limx→∞

(lnx)k+1

x

�= limx→∞

(k + 1)(lnx)k

x= 0 (since k ∈ S).

Thus k + 1 ∈ S, and S is the set of positive integers.

(b) Choose any positive number α. Choose a positive integer k > α.

Then, for x > e,

0 <(lnx)α

x<

(lnx)k

x

and the result follows by the pinching theorem and part (a).

59. Let y =(a1/x + b1/x

2

)x

. Then ln y = x ln[a1/x + b1/x

2

]=

ln(a1/x + b1/x

2

)1/x

.

Now,

limx→∞

ln(a1/x + b1/x

2

)1/x

�= limx→∞

2a1/x + b1/x

· −a1/x ln a− b1/x ln b

2x2

−1/x2

= limx→∞

a1/x ln a + b1/x ln b

a1/x + b1/x= 1

2 ln ab = ln√ab

Thus, limx→∞

ln y = ln√ab =⇒ lim

x→∞y =

√ab.

60. (a) limk→0+

v(t) = limk→0+

mg(1 − e−(k/m)t

)k

�= limk→0+

gte−(k/m)t

1= gt

(b)dv

dt= g =⇒ v(t) = gt + C; v(0) = 0 =⇒ C = 0 and v(t) = gt.

61. (a) Ab = 1 − (1 + b)e−b

(b) xb =2 −

(2 + 2b + b2

)e−b

1 − (1 + b)e−b; yb =

14 − 1

4

(1 + 2b + 2b2

)e−2b

2 [1 − (1 + b)e−b]

(c) limb→∞

Ab = 1; limb→∞

xb = 2; limb→∞

yb =18

62. (a) Vx = π

[14− 1

4(1 + 2b + 2b2

)e−2b

]

(b) Vy = 2π[2 −

(2 + 2b + b2

)e−b]

(c) limb→∞

Vx =π

4, lim

b→∞Vy = 4π



616 SECTION 11.6

63. (a) limx→0+

(1 + x2

)1/x = 1

1 2x

1

2

3

y

f

g

(b) limx→0+

(1 + x2

)1/x = 1 since

limx→0+

ln[(

1 + x2)1/x] = lim

x→0+

ln(1 + x2

)x

�= limx→0+

2x(1 + x2)

= 0

64. (a) limx→∞

f(x) ∼= 1.5

5 10 15 20x

1.5

y

(b) limx→∞

f(x) = limx→∞

[√x2 + 3x + 1 − x

]= lim

x→∞3x + 1√

x2 + 3x + 1 + x= lim

x→∞3x + 1

x√

1 + (3/x) + 1/x2 + x=

32

65. (a) limx→∞

g(x) ∼= − 1.7.10 50

x

-2

y

g

(b) limx→∞

g(x) = limx→∞

[3√x3 − 5x2 + 2x + 1 − x

]= lim

x→∞−5x2 + 2x + 1(

3√x3 − 5x2 + 2x + 1

)2+ x 3

√x3 − 5x2 + 2x + 1 + x2

= − 53∼= −1.667

66. [P (x)]1/n − x =([P (x)]1/n − x

)· [P (x)](n−1)/n + x[P (x)](n−2)/n + · · · + xn−2[P (x)]1/n + xn−1

[P (x)](n−1)/n + x[P (x)](n−2)/n + · · · + xn−2[P (x)]1/n + xn−1

=P (x) − xn

[P (x)](n−1)/n + x[P (x)](n−2)/n + · · · + xn−2[P (x)]1/n + xn−1

=b1x

n−1 + b2xn−2 + · · · + bn

[P (x)](n−1)/n + x[P (x)](n−2)/n + · · · + xn−2[P (x)]1/n + xn−1→ b1

nas x → ∞



SECTION 11.7 617

SECTION 11.7

1. 1 :∫ ∞

1

dx

x2= lim

b→∞

∫ b

1

dx

x2= lim

b→∞

[− 1x

]b1

= limb→∞

[1 − 1

b

]= 1

2.π

2:

∫ ∞

0

dx

1 + x2= lim

b→∞tan−1 b =

π

2

3.π

4:

∫ ∞

0

dx

4 + x2= lim

b→∞

∫ b

0

dx

4 + x2= lim

b→∞

[12

tan−1 x

2

]b0

= limb→∞

12

tan−1

(b

2

)=

π

4

4.1p

:∫ ∞

0

e−px dx = limb→0

(−e−pb

p+

1p

)=

1p

5. diverges:∫ ∞

0

epxdx = limb→∞

∫ b

0

epx dx = limb→∞

[1pepx]b0

= limb→∞

1p

(epb − 1

)= ∞

6. 2 :∫ 1

0

dx√x

= lima→0+

∫ 1

a

dx√x

= lima→0+

(2 − 2√a) = 2

7. 6:∫ 8

0

dx

x2/3= lim

a→0+

∫ 8

a

x−2/3 dx = lima→0+

[3x1/3

]8a

= lima→0+

[6 − 3a1/3

]= 6

8. diverges:∫ 1

0

dx

x2= lim

a→0+

∫ 1

a

dx

x2= lim

a→0+

(−1 +

1a

)= ∞

9.π

2:

∫ 1

0

dx√1 − x2

= limb→1−

∫ b

0

dx√1 − x2

= limb→1−

sin−1 b =π

2

10. 2:∫ 1

0

dx√1 − x

= lima→0+

∫ 1

a

dx√1 − x

= lima→0+

[−2

√1 − x

]1a

= lima→0+

(0 + 2√

1 − a) = 2

11. 2:∫ 2

0

x√4 − x2

dx = limb→2−

∫ b

0

x(4 − x2

)−1/2dx = lim

b→2−

[−(4 − x2

)1/2]b0

= limb→2−

(2 −

√4 − b2

)= 2

12.π

2:

∫ a

0

dx√a2 − x2

= limb→a−

∫ b

0

dx√a2 − x2

= limb→a−

[sin−1

(b

a

)− sin−1(0)

]=

π

2

13. diverges:∫ ∞

e

lnx

xdx = lim

b→∞

∫ b

e

lnx

xdx = lim

b→∞

[12

(lnx)2]be

= limb→∞

[12

(ln b)2 − 12

]= ∞


e

dx

x lnx= lim

b→∞

∫ b

e

dx

x lnx= lim

b→∞[ln(lnx)]be = ∞



618 SECTION 11.7

15. −14

: ∫ 1

0

x lnx dx = lima→0+

∫ 1

a

x lnx dx = lima→0+

[12x2 lnx− 1

4x2

]1

a

(by parts)∧

= lima→0+

[14a2 − 1

2a2 ln a− 1

4

]= −1

4

Note: limt→0+

t2 ln t = limt→0+

ln t

1/t2�= lim

t→0+

1/t−2/t3

= −12

limt→0+

t2 = 0.

16. 1:∫ ∞

e

dx

x(lnx)2= lim

b→∞

∫ b

e

dx

x(lnx)2= lim

b→∞

[− 1

lnx

]be

= limb→∞

[− 1

ln b+ 1]

= 1

17. π:∫ ∞

−∞

dx

1 + x2= lim

a→−∞

∫ 0

a

dx

1 + x2+ lim

b→∞

∫ b

0

dx

1 + x2

= lima→−∞

[tan−1 x

]0a

+ limb→∞

[tan−1 x

]b0

= −(−π

2

)+

π

2= π

18.ln 32

:∫ ∞

2

dx

x2 − 1= lim

b→∞

∫ b

2

dx

x2 − 1= lim

b→∞

[12

ln∣∣∣∣x− 1x + 1

∣∣∣∣]b2

= limb→∞

[12

ln(b− 1b + 1

)+

12

ln 3]

=12

ln 3


−∞

dx

x2= lim

a→−∞

∫ −1

a

dx

x2+ lim

b→0−

∫ b

−1

dx

x2+ lim

c→0+

∫ 1

c

dx

x2+ lim

d→∞

∫ d

1

dx

x2;

and, limc→0+

∫ 1

c

dx

x2= lim

c→0+

[− 1x

]1

c

= limc→0+

[1c− 1]

= ∞

20. 2:∫ 3

1/3

dx3√

3x− 1= lim

a→ 13+

∫ 3

a

dx

(3x− 1)1/3= lim

a→ 13+

[3(3x− 1)2/3

2 · 3

]3

a

= lima→ 1

3+

[82/3

2− (3a− 1)2/3

2

]= 2

21. ln 2: ∫ ∞

1

dx

x(x + 1)= lim

b→∞

∫ b

1

[1x− 1

x + 1

]dx

= limb→∞

[ln(

x

x + 1

)]b1

= limb→∞

[ln(

b

b + 1

)− ln

(12

)]= ln 2

22. −1 :∫ 0

−∞xex dx = lim

a→−∞

∫ 0

a

xex dx = lima→−∞

[xex − ex]0a = lima→−∞

[−1 − aea + ea] = −1

23. 4:

∫ 5

3

x√x2 − 9

dx = lima→3−

∫ 5

a

x(x2 − 9

)−1/2dx

= lima→3−

[(x2 − 9

)1/2]5a

= lima→3−

[4 −

(a2 − 9

)1/2] = 4



SECTION 11.7 619

24. ∫ 4

1

dx

x2 − 4= lim

b→2−

∫ b

1

dx

x2 − 4+ lim

a→2+

∫ 4

a

dx

x2 − 4= lim

b→2−

[14

ln∣∣∣∣x− 2x + 2

∣∣∣∣]b1

+ lima→2+

[14

ln∣∣∣∣x− 2x + 2

∣∣∣∣]4

a

= limb→2−

(14

ln∣∣∣∣b− 2b + 2

∣∣∣∣− 14

ln13

)+ lim

a→2+

(14

ln26− 1

4ln∣∣∣∣a− 2a + 2

∣∣∣∣)

= ∞ diverges

25.∫ 3

−3

dx

x(x + 1)diverges since

∫ 3

0

dx

x(x + 1)diverges:

∫ 3

0

dx

x(x + 1)= lim

a→0+

∫ 3

a

(1x− 1

x + 1

)dx = lim

a→0+

[ln |x| − ln |x + 1|

]3a

= lima→0+

[ ln 3 − ln 4 − ln a + ln (a + 1)] = ∞.

26.14

:∫ ∞

1

x

(1 + x2)2dx = lim

b→∞

∫ b

1

x

(1 + x2)2dx = lim

b→∞

[ −12(1 + x2)

]b1

= limb→∞

[ −12(1 + b2)

+14

]=

14

27.∫ 1

−3

dx

x2 − 4diverges since

∫ 1

−2

dx

x2 − 4diverges:

∫ 1

−2

dx

x2 − 4= lim

a→−2+

∫ 1

a

14

[1

x− 2− 1

x + 2

]dx

= lima→−2+

[14

(ln |x− 2| − ln |x + 2|)]1

a

= lima→−2+

14

[− ln 3 − ln |a− 2| + ln |a + 2| ] = −∞.

28.π

2: ∫ ∞

0

1ex + e−x

dx = limb→∞

∫ b

0

1ex + e−x

dx = limb→∞

[tan−1 ex

]b0

=π

2− π

4

∫ 0

−∞

1ex + e−x

dx = limb→−∞

∫ 0

b

1ex + e−x

dx = limb→−∞

[tan−1 ex

]0b

=π

4


0

coshx dx = limb→∞

∫ b

0

coshx dx = limb→∞

[sinhx]b0 = ∞

30. Since∫ 2

1

dx

x2 − 5x + 6= lim

b→2−

∫ b

1

dx

(x− 2)(x− 3)= lim

b→2−

[ln∣∣∣∣x− 3x− 2

∣∣∣∣]b1

= limb→2−

(ln∣∣∣∣b− 3b− 2

∣∣∣∣− ln 2)

diverges, so does∫ 4

1

dx

x2 − 5x + 6

31.12

:∫ ∞

0

e−x sinx dx = limb→∞

∫ b

0

e−x sinx dx = limb→∞

− 12[e−x cosx + e−x sinx

]b0

(by parts)∧

= limb→∞

12[1 − e−b cos b− e−b sin b

]=

12



620 SECTION 11.7


0

cos2 x dx = limb→∞

[x

2+

sin 2x4

]b0

= limb→∞

(b

2+

sin 2b4

)= ∞

33. 2e− 2 :∫ 1

0

e√x

√xdx = lim

a→0+

∫ 1

a

e√x

√xdx = lim

a→0+

[2 e

√x]1a

= 2(e− 1)

34. 2 :∫ π/2

0

cosx√sinx

dx = lima→0+

∫ π/2

a

cosx√sinx

dx = lima→0+

[2√

sinx]π/2a

= 2

35. (a) converges:∫ ∞

0

x

(16 + x2)2dx =

132

(b) converges:∫ ∞

0

x2

(16 + x2)2dx =

π

16

(c) converges:∫ ∞

0

x

16 + x4dx =

π

16(d) diverges

36. (a) converges:∫ 2

0

x3

3√

2 − xdx =

243 3√

455

(b) converges:∫ 2

0

1√2 − x

dx = 2√

2

(c) converges:∫ 2

0

x√2 − x

dx =8√

23

(d) converges:∫ 2

0

1√2x− x2

dx = π

37.∫ 1

0

sin−1 x dx =[x sin−1 x

]10−∫ 1

0

x√1 − x2

dx =π

2− lim

a→1−

∫ a

0

x√1 − x2

dx

(by parts)

Now,∫ a

0

x√1 − x2

dx = − 12

∫ 1−a2

1

1√udu =

[−√u]1−a2

1= 1 −

√1 − a2

u = 1 − x2

Thus,∫ 1

0

sin−1 x dx =π

2− lim

a→1−

(1 −

√1 − a2

)=

π

2− 1.

38. (a)∫ ∞

0

xre−x dx diverges if r ≤ −1:

∫ ∞

0

xre−x dx =∫ 1

0

xre−xdx +∫ ∞

1

xre−x dx and∫ 1

0

xre−x dx diverges.

For any r > −1, we can find k such that xr < ex/2 for x ≥ k (ex/2 grows faster than any power

of x). Then∫ ∞

0

xre−xdx <

∫ k

0

xre−xdx +∫ ∞

k

e−x/2dx, which converges. Thus∫ ∞

0

xre−xdx

converges for all r > −1.

(b) For n = 1 :∫ ∞

0

xe−x dx = limb→∞

[−xe−x − e−x

]b0

= limb→∞

[−be−b − e−b + 1

]= 1

Assume true for n.∫ ∞

0

xn+1e−x dx = limb→∞

([−xn+1e−x

]b0

+ (n + 1)∫ b

0

xne−x dx

)

= limb→∞

(−bn+1e−b) + (n + 1)∫ ∞

0

xne−x dx = 0 + (n + 1)n! = (n + 1)!



SECTION 11.7 621

39.

∫ ∞

0

1√x (1 + x)

dx =∫ 1

0

1√x (1 + x)

dx +∫ ∞

1

1√x (1 + x)

dx

= lima→0+

∫ 1

a

1√x (1 + x)

dx + limb→∞

∫ b

1

1√x (1 + x)

dx

Now,∫

1√x (1 + x)

dx =∫

21 + u2

du = 2 arctanu + C = 2 arctan√x + C.

u =√x

Therefore, lima→0+

∫ 1

a

1√x (1 + x)

dx = lima→0+

[2 arctan

√x]1a

= lima→0+

2[π/4 − arctan

√a)] =

π

2

and limb→∞

∫ b

1

1√x (1 + x)

dx = limb→∞

[2 arctan

√x]b1

= limb→∞

2[arctan

√b− π/4

]=

π

2.

Thus,∫ ∞

0

1√x (1 + x)

dx = π.

40.∫ ∞

1

1x√x2 − 1

dx = lima→1+

∫ 2

a

1x√x2 − 1

dx + limb→∞

∫ b

2

1x√x2 − 1

dx

= lima→1+

[sec−1 x

]2a

+ limb→∞

[sec−1 x

]∞2

= lima→1+

(sec−1 2 − sec−1 a) + limb→∞

(sec−1 b− sec−1 2)

= (sec−1 2 − 0) + (π

2− sec−1 2) =

π

2

41. surface area S =∫ ∞

1

2π(

1x

)√1 +

1x4

dx = 2π∫ ∞

1

√x4 + 1x3

dx = ∞ by comparison with∫ ∞

1

1xdx

42. A =∫ π/2

0

(secx− tanx) dx = limb→π/2−

∫ b

0

(secx− tanx) dx = limb→π/2−

[ln(secx− tanx) − ln secx

]b0

= limb→π/2−

[ln(1 + sinx)

]b0

= ln 2

43. (a) (b) A =∫ 1

0

1√xdx = lim

a→0+

∫ 1

a

1√xdx

= lima→0+

[2√x]1a

= 2

(c) V =∫ 1

0

π

(1√x

)2

dx = π

∫ 1

0

1xdx = π lim

a→0+

∫ 1

a

1xdx = π lim

a→0+[lnx]1a diverges



622 SECTION 11.7

44. (a) (b) A =∫ ∞

0

11 + x2

dx = limb→∞

tan−1 b =π

2

(c) Vx =∫ ∞

0

π · 1(1 + x2)2

dx = limb→∞

π

2

[tan−1 x +

x

1 + x2

]b0

= limb→∞

π

2

(tan−1 b +

b

1 + b2− 0)

=π2

4

(d) Vy =∫ ∞

0

2πx1 + x2

dx = limb→∞

π[ln(1 + x2)

]b0

= ∞

45. (a) (b) A =∫ ∞

0

e−x dx = 1

(c) Vx =∫ ∞

0

πe−2x dx = π/2

(d) Vy =∫ ∞

0

2πxe−x dx = limb→∞

∫ b

0

2πxe−x dx = limb→∞

[2π(−x− 1)e−x

]b0

(by parts)

= 2π(

1 − limb→∞

b + 1eb

)= 2π(1 − 0) = 2π

(e) A =∫ ∞

0

2πe−x√

1 + e−2x dx = limb→∞

∫ b

0

2πe−x√

1 + e−2x dx

∫ b

0

2πe−x√

1 + e−2x dx = −2π∫ e−b

1

√1 + u2 du

u = e−x

= −π[u√

1 + u2 + ln(u +

√1 + u2

)]e−b

1

= π[√

2 + ln(1 +

√2)− e−b

√1 + e−2b − ln

(e−b +

√1 + e−2b

)]Taking the limit of this last expression as b → ∞, we have

A = π[√

2 + ln(1 +

√2)]

.

46. xA =∫ ∞

0

xe−x dx = 1, x =xA

A= 1

yA =∫ ∞

0

12e−2x dx =

14, y =

yA

A=

14; centroid: (1, 1

4 )

Yes: 2πxA = 2π = Vy, 2πyA =12π = Vx



SECTION 11.7 623

47. (a) The interval [0, 1] causes no problem. For x ≥ 1, e−x2 ≤ e−x and∫ ∞

1

e−x dx is finite.

(b) Vy =∫ ∞

0

2πxe−x2dx = lim

b→∞

∫ b

0

2πxe−x2dx = lim

b→∞π[−e−x2

]b0

= limb→∞

π(1 − e−b2

)= π

48. (a) A =∫ ∞

1

(1x− x

x2 + 1

)dx =

12

ln 2

(b) Vx =∫ ∞

1

[(1x

)2

−(

x

x2 + 1

)2]dx <

∫ ∞

1

dx

x2finite

(c) Vy =∫ ∞

1

2πx(

1x− x

x2 + 1

)dx = 2π

∫ ∞

1

1x2 + 1

dx =12π2

49. (a) (b) A = lima→0+

∫ 1

a

x−1/4 dx = lima→0+

[43x3/4

]1

a

=43

(c) Vx = lima→0+

∫ 1

a

πx−1/2 dx = lima→0+

[2πx1/2

]1a

= 2π

(d) Vy = lima→0+

∫ 1

a

2πx3/4 dx = lima→0+

[8π7x7/4

]1

a

=87π

50. (i) Suppose that∫ ∞

a

g(x) dx = L. Since f(x) ≥ 0 for x ∈ [a,∞),∫ x

a

f(t) dt is increasing.

Therefore it is sufficient to show that∫ x

a

f(t) dt is bounded above. For any number

M ≥ a, we have ∫ M

a

f(x) dx ≤∫ M

a

g(x) dx ≤∫ ∞

a

g(x) dx = L

Therefore,∫ x

a

f(t) dt is bounded and∫ ∞

a

f(x) dx converges.

(ii) If∫ ∞

0

f(x) dx diverges, then∫ ∞

0

g(x) dx can not converge, by (i)

51. converges by comparison with∫ ∞

1

dx

x3/252. converges by comparison with

∫ ∞

2

e−x dx on [2,∞).

53. diverges since for x large the integrand is greater than1x

and∫ ∞

1

dx

xdiverges

54. Converges by comparison with∫ ∞

π

dx

x255. converges by comparison with

∫ ∞

1

dx

x3/2

56. Diverges by comparison with∫ ∞

e

dx

(x + 1) ln(x + 1)



624 SECTION 11.7

57. (a) limb→∞

∫ b

0

2x1 + x2

dx = limb→∞

[ln(1 + x2

) ]b0

= ∞

Thus, the improper integral∫ ∞

0

2x1 + x2

dx diverges.

(b)limb→∞

∫ b

−b

2x1 + x2

dx = limb→∞

[ln(1 + x2

) ]b−b

= limb→∞

(ln[1 + b2

]− ln

[1 + (−b)2

])= lim

b→∞(0) = 0

58.∫ ∞

0

sinx dx = limb→∞

∫ b

0


[− cosx

]b0

= 1 − limb→∞

cos b; the limit does not exist

limb→∞

∫ b

−b


[− cosx

]b−b

= limb→∞

[− cos b + cos(−b)] = limb→∞

[ 0 ] = 0

59. r(θ) = aecθ, r′(θ) = acecθ

L =∫ θ1

−∞

√a2e2cθ + a2c2e2cθ dθ

(10.7.3)∧

=(a√

1 + c2)(

limb→−∞

∫ θ1

b

ecθ dθ

)

=(a√

1 + c2)(

limb→−∞

[ecθ

c

]θ1b

)

=

(a√

1 + c2

c

)(lim

b→−∞

[ecθ1 − ecb

])=

(a√

1 + c2

c

)ecθ1

60. For all real t, − t2

2< t + 1. Therefore

∫ x

−∞e−t2/2 dt converges by comparison with

∫ x

−∞et+1 dt.

61. F (s) =∫ ∞

0

e−sx · 1 dx = limb→∞

∫ b

0

e−sx dx = limb→∞

[− 1

se−sx

]b0

=1s

provided s > 0.

Thus, F (s) =1s; dom(F ) = (0,∞).

62. F (s) =∫ ∞

0

xe−sx dx = limb→∞

[−xe−sx

s− e−sx

s2

]b0

= limb→∞

(−be−sb

s− e−sb

s2+

1s2

)

=1s2

if s > 0, diverges for s ≤ 0, so dom(F ) = (0,∞).

63. F (s) =∫ ∞

0

e−sx cos 2x dx = limb→∞

∫ b

0

e−sx cos 2x dx

Using integration by parts∫

e−sx cos 2x dx =4

s2 + 4

[12e−sx sin 2x− s

4e−sx cos 2x

]+ C.



SECTION 11.7 625

Therefore,

F (s) = limb→∞

4s2 + 4

[12e−sx sin 2x− s

4e−sx cos 2x

]b0

=4

s2 + 4limb→∞

[12e−sb sin 2b− s

4e−sb cos 2b +

s

4

]=

4s2 + 4

· s4

=s

s2 + 4provided s > 0.

Thus, F (s) =s

s2 + 4; dom(F ) = (0,∞).

64. F (s) =∫ ∞

0

eaxe−sx dx =∫ ∞

0

e(a−s)x dx = limb→∞

[e(a−s)b

a− s− 1

a− s

]

=1

s− aif s > a, diverges if s ≤ a; so dom F = (a,∞)

65. The function f is nonnegative on (−∞,∞) and∫ ∞

−∞f(x) dx =

∫ 0

−∞0 dx +

∫ ∞

0

6x(1 + 3x2)2

dx =∫ ∞

0

6x(1 + 3x2)2

dx

Now,∫

6x(1 + 3x2)2

dx = − 11 + 3x2

+ C.

Therefore, ∫ ∞

−∞f(x) dx = lim

b→∞

[− 1

1 + 3x2

]b0

= limb→∞

(1 − 1

1 + 3b2

)= 1.

66. f is nonnegative, and∫ ∞

−∞f(x) dx =

∫ ∞

0

ke−kx dx = limb→∞

[−e−kx

]b0

= limb→∞

(−e−kb + 1

)= 1

So, f is a probability density function.

67. μ =∫ ∞

−∞xf(x) dx =

∫ 0

−∞0 dx +

∫ ∞

0

kxe−kx dx = limb→∞

∫ b

0

kxe−kx dx

Using integration by parts,∫

kxe−kx dx = −xe−kx − 1ke−kx + C.

Therefore,

μ =∫ ∞

−∞xf(x) dx = lim

b→∞

[−xe−kx − 1

ke−kx

]b0

= limb→∞

[−be−kb − 1

ke−kb +

1k

]=

1k

68. σ =∫ ∞

−∞(x− μ)2f(x) dx =

∫ ∞

0

(x− 1

k

)2

ke−kx dx

=∫ ∞

0

kx2e−kx dx− 2∫ ∞

0

xe−kx dx +1k

∫ ∞

0

e−kx dx =1k2

69. Observe that F (t) =∫ t

1

f(x) dx is continuous and increasing, that an =∫ n

1

f(x) dx

is increasing, and that (∗) an ≤∫ t

1 f(x) dx ≤ an+1 for t ∈ [n, n + 1 ].



626 REVIEW EXERCISES

If∫ ∞

1

f(x) dx converges, then F , being continuous, is bounded and, by (∗), {an} is bounded

and therefore convergent. If {an} converges, then {an} is bounded and, by (∗), F is bounded. Being

increasing, F is also convergent; i.e.,∫ ∞

1

f(x) dx converges.

REVIEW EXERCISES

1. |x− 2| ≤ 3 =⇒ −1 ≤ x ≤ 5 : lub = 5, glb = −1.

2. x2 > 3 =⇒ x >√

3 or x < −√

3; no lub, no glb.

3. x2 − x− 2 ≤ 0 =⇒ (x− 2)(x + 1) ≤ 0 =⇒ −1 ≤ x ≤ 2 : lub = 2, glb = −1.

4. cosx ≤ 1 for all x; no lub, no glb.

5. Since e−x2 ≤ 1 for all x, e−x2 ≤ 2 for all x; no lub, no glb.

6. lnx < e =⇒ 0 < x < ee : lub = ee, glb = 0.

7. increasing; bounded below by 12 and above by 2

3 .

8. increasing; bounded below by 0 but not bounded above:n2 − 1

n= n− 1

n→ ∞ as n → ∞.

9. bounded below by 0 and above by 32 ; not monotonic

10. increasing; bounded below by 45 and above by 1.

11.{

2n

n2

}={2, 1, 8

9 , 1, 3225 , . . .

}; the sequence is not monotonic.

However, it is increasing from a3 on. The sequence is bounded below by 89 ; it is not bounded above.

12.{

sin (nπ/2)n2

}={1, 0, − 1

9 , 0, 125 , · · ·

}; bounded below by − 1

9 and above by 1; not monotonic

13. the sequence does not converge; n21/n → ∞ as n → ∞

14. converges to 1:n2 + 3n + 2n2 + 7n + 12

=1 +

3n

+2n2

1 +7n

+12n2

→ 1.

15. converges to 1: limn→∞

1n

ln(

n

n + 1

)= 0 =⇒ lim

n→∞

(n

1 + n

)1/n

= 1



REVIEW EXERCISES 627

16. converges to 0:4n2 + 5n + 1

n3 + 1=

4n

+5n2

+1n3

1 +1n3

→ 0.

17. converges to 0: cos πn sin π

n → cos 0 sin 0 = 0 as n → ∞

18. diverges: (2 + 1n )n > 2n and 2n diverges.

19. converges to 0: 0 = [ln 1]n ≤ [ln(1 +1n

) ]n ≤ [ln 2]n; [ln 2]n → 0 as n → ∞.

20. converges to ln 8: 3 ln 2n− ln(n3 + 1) = ln8n3

n3 + 1→ ln 8.

21. converges to 32 :

3n2 − 1√4n4 + 2n2 + 3

=3 − 1

n2√4 + 2

n2 + 3n4

→ 32

as n → ∞.

22. converges to 0:(n2 + 4)

13

2n + 1=

(1/n + 4/n3)1/3

2 + 1/n→ 0 as n → ∞.

23. converges to 0:π

ncos

π

n→ 0 cos 0 = 0 as n → ∞.

24. converges 0: (n/π) sin(nπ) = 0 for all positive integers n.

25. converges to 0:∫ n+1

n

e−xdx =[− e−x

]n+1

n= e−n(1 − 1

e) → 0 as n → ∞

26. diverges:∫ n

1

1√xdx =

[2√x]n1

= 2√n− 2 and 2

√n− 2 diverges.

27. Given ε > 0. Since an → L, there exists a positive integer K such that if n ≥ K, then |an − L| < ε.

Now, if n ≥ K − 1, then n + 1 ≥ K and |an+1 − L| < ε. Therefore, an+1 → L.

28. Let ε > 0. Since an → L, there is positive integer K such that if n ≥ K,

|an − L| < ε

2.

The set {|a1 − L|, · · · , |aK − L|} is a finite set so there is a positive integer N such that if n > N ,

|ai − L|n

<ε

2K, i = 1, 2, · · · ,K.

Let M = max {K,N}. Then, if n ≥ M ,∣∣∣∣a1 + · · · + ann

− L

∣∣∣∣ ≤ K∑i=1

|ai − L|n

+n∑

j=K+1

|aj − L|n

< K(ε

2K) + n(

ε

2n) = ε.

Therefore mn → L.




29. As an example, let a = π3 . Then

cos π/3 = 0.5, cos cos 0.5 ∼= 0.87758, · · · .

Using technology (graphing calculator, CAS), we get

cos cos · · · cos π/3 → 0.73910.

and cos (0.73910) ∼= 0.73910.

Hence, numerically, this sequence converges to 0.73910.

30. Let f(x) = sin (cosx) and let a = π/3. Then

f(π/3) ∼= 0.4794, f(f(π/3)) ∼= 0.7753, · · ·

After 14 steps, we get f(f(· · · f(π/3)) ∼= 0.6948 and sin (cos 0.6948) ∼= 0.6948.

31. limx→∞

5x + 2 lnx

x + 3 lnx= lim

x→∞

5 + 2lnx

x

1 + 3lnx

x

= 5;(

limx→∞

lnx

x= 0

)

32. limx→0

ex − 1tan 2x

�= limx→0

ex

2 sec2 2x=

12

33. limx→0

ln(cosx)x2

�= limx→0

− sinx

cosx2x

= limx→0

−12 cosx

· sinx

x= −1

2

34. Set y = x1/(x−1). Then ln y =ln x

x− 1, and lim

x→1

ln x

x− 1�= lim

x→1

1x

= 1

Therefore, limx→1

x1/(x−1) = e.

35. limx→∞

(1 +4x

)2x = limx→∞

[(1 +

4x

)x]2

= e8

36. limx→0

e2x − e−2x

sinx

�= limx→0

2e2x + 2e−2x

cosx= 4

37. limx→0+

x2 lnx = limx→0+

lnx1x2

�= limx→0+

1x−2x3

= limx→0+

−x2

2= 0

38. limx→∞

10x

x10

�= limx→∞

10x ln 1010x9

�= limx→∞

10x(ln 10)2

(10)(9)x8

�= · · · = limx→∞

10x(ln 10)10

10!→ ∞

39. limx→0

ex + e−x − x2 − 2sin2 x− x2

�= limx→0

ex − e−x − 2x2 sinx cosx− 2x

�= limx→0

ex − e−x − 2sin 2x− 2x

�= limx→0

ex + e−x − 22 cos 2x− 2

�= limx→0

ex − e−x

−4 sinx2x�= lim

x→0

ex + e−x

−8 cos 2x= −1

4




40. limx→1

csc(πx) lnx = limx→1

lnx

sinπx

�= limx→1

1/xπ cosπx

= − 1π

41. limx→∞

xe−x2∫ x

0

et2dt = lim

x→∞

∫ x

0

et2dt

ex2

x

�= limx→∞

ex2

2x2ex2 − ex

2

x2

= limx→∞

x2

2x2 − 1=

12

42. Consider lne−1/x2

xn=(− 1x2

− n ln |x|)

.

limx→0

(− 1x2

− n ln |x|)

= limx→0

−1 + nx2 ln |x|x2

= −∞

Therefore limx→0

e−1/x2

xn→ 0.

43.∫ ∞

1

e−√x

√x

dx = limb→∞

∫ b

1

e−√x

√x

dx = limb→∞

[− 2e−

√x]b1

= limb→∞

(−2e−√b + 2e−1) = 2e−1

44.∫

x√1 − x2

dx = −√

1 − x2 + C

∫ 1

0

x√1 − x2

dx = limb→1−

∫ b

0

x√1 − x2

dx = limb→1−

[−√

1 − x2]b0

= limb→1−

(−√

1 − b2 + 1) = 1

45.∫ 1

0

11 − x2

dx = lima→1−

∫ a

0

11 − x2

dx = −12

lima→1−

∫ a

0

(1

x− 1− 1

x + 1

)dx (partial fractions)

=12

lima→1−

[ln(x + 1) − ln(1 − x)

]a0

=12

lima→1−

ln[a + 11 − a

]= ∞;

the integral diverges.

46.∫ π/2

0

sec x dx = limc→π/2−

∫ c

0

sec x dx = limc→π/2−

[ln(secx + tanx)

]c0

= limc→π/2−

ln(sec c + tan c) = ∞;

the integral diverges.

47.∫ ∞

1

sin(π/x)x2

dx = limb→∞

∫ b

1

sin(π/x)x2

dx = limb→∞

[ 1π

cosπ/x]b1

=2π

48.∫ 9

0

1(x− 1)2/3

dx = limc→1−

∫ c

0

1(x− 1)2/3

dx + limc→1+

∫ 9

c

1(x− 1)2/3

dx

limc→1−

∫ c

0

1(x− 1)2/3

dx = limc→1−

3[(x− 1)1/3

]c0

= 3

limc→1+

∫ 9

c

1(x− 1)2/3

dx = limc→1+

3[(x− 1)1/3

]9c

= 6

∫ 9

0

1(x− 1)2/3

= 3 + 6 = 9.




49.∫

1ex + e−x

dx =∫

ex

e2x + 1dx = arctan ex + C

∫ ∞

0

1ex + e−x

dx = limc→∞

∫ c

0

1ex + e−x

dx = limc→∞

arctan ex∣∣∣c0

=π

2

50. Set u = ln x, du =1xdx; u(2) = ln 2. Then∫ ∞

2

1x(lnx)k

dx =∫ ∞

ln 2

1uk

du

The integral converges if k > 1 and diverges otherwise.

For k > 1, ∫ ∞

2

1x(lnx)k

dx = limc→∞

∫ c

ln 2

1uk

du = limc→∞

11 − k

u1−k∣∣∣cln 2

=1

(k − 1) (ln 2)k−1

51.∫ a

0

ln(1/x) dx =∫ a

0

− ln x dx = limc→0+

∫ a

c

− ln x dx = limc→0+

[− x lnx + x

]ac

= limc→0+

[−a ln a + a + c ln c− c] = a ln(1/a) + a

52. y = (a2/3 − x2/3)3/2; y′ = −x−1/3(a2/3 − x2/3)1/2

L =∫ a

0

√1 + (y′)2 dx =

∫ a

0

a1/3x−1/3 dx =3a2

53. For any a ∈ S + T , a = s + t for some s ∈ S and t ∈ T . Hence a ≤ lub(S) + lub(T ).

Therefore, S + T is bounded above and lub(S) + lub(T ) is an upper bound for S + T .

Let M = lub(S + T ) and suppose M < lub(S) + lub(T ). Set ε = (lub(S) + lub(T )) −M . There

exist s ∈ S and t ∈ T such that

lub(S) − s < ε/2, and lub(T ) − t < ε/2

Now,

lub(S) + lub(T ) − (s + t) = lub(S) − s + lub(T ) − t < ε = lub(S) + lub(T ) −M

which implies s + t > M , a contradiction. Therefore lub(S) + lub(T ) = lub(S + T ).

54. (a) Since S is bounded below, there is a number b such that b ≤ s for every s ∈ S. Thus b is a

lower bound for S and B �= ∅.(b) Choose any s ∈ S. Then, for any b ∈ B, b ≤ s. Therefore B is bounded above (each element

s ∈ S is an upper bound for B).

(c) We show first that glb(S) is an upper bound for B. For if not, there is b ∈ B such that b > glb(S).

Then there is an s ∈ S such that glb(S) < s < b, which contradicts the fact that b is a lower

bound of S. It follows that lub(B) ≤ glb(S). If lub(B) < glb(S), then there exists a number

a such that lub(B) < a < glb(S) which implies that a is a lower bound for S and a ∈ B.

Therefore a ≤ lub(B), a contradiction. Thus, lub(B) = glb(S).




55. (a) If∫ ∞

−∞f(x)dx = L, then

limc→∞

∫ c

0

f(x) dx, and limb→−∞

∫ 0

b

f(x) dx

both exist and

limc→∞

∫ c

0

f(x) dx + limb→−∞

∫ 0

b

f(x) dx = L

Let c = −b, then

limc→∞

∫ c

−c

f(x) dx = L

(b) Set f(x) = x. then limc→∞

∫ c

−c

x dx = 0, but∫ ∞

−∞x dx diverges.

56. (a) Assume that f is nonnegative on (−∞,∞).

By Exercise 55,∫ ∞

−∞f(x) dx = L =⇒ lim

c→∞

∫ c

−c

f(x) dx = L

Now assume that limc→∞

∫ c

−c

f(x) dx = L. Since f is nonnegative,

∫ x

0

f(t) dt ≤ L on [0,∞).

Therefore∫ x

0

f(t) dt is a bounded and nondecreasing function, which implies that

limc→∞

∫ c

0

f(x) dx exists. Similarly, limc→∞

∫ 0

−c

f(x) dx exists.

Therefore,∫ ∞

−∞f(x) dx exists, and, by the uniqueness of the limit,

∫ ∞

−∞f(x) dx = L.

57. Let S be a set of integers which is bounded above. Then there is an integer k ∈ S such that k ≥ n

for all n ∈ S, for if not, S is not bounded above. Therefore, k is an upper bound for S.

Let M = lub(S). Then M ≥ k since k ∈ S. Also M ≤ k since k is an upper bound for S. Therefore

M = k; the least upper bound of S is an element of S.

58. lub [Lf (P )] =∫ b

a f(x) dx; glb [Uf (P )] =∫ b

a f(x) dx.

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Calculus one and several variables 10E Salas solutions manual ch11

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