Calculus one and several variables 10E Salas solutions manual ch11

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SECTION 11.1

CHAPTER 11 SECTION 11.1 1.

lub = 2;

3.

no lub;

5.

lub = 2;

7.

no lub;

9.

lub = 2 12 ;

glb = 0

2.

lub = 2;

glb = 0

4.

lub = 1,

no glb

6.

lub = 3;

glb = −1

8.

lub = 2;

glb = −2

glb = 2

10.

lub = 0;

glb = −1

glb = 0 glb = −2 glb = 2

glb = −2 19

11.

lub = 1;

glb = 0.9

12.

lub = 2 19 ,

13.

lub = e;

glb = 0

14.

no lub,

glb = 1

15.

lub = 12 (−1 +

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,

22.

glb = 1;

1 ≤ 1 < 1 + 0.0001

√

5);

0≤



 23. glb S = 0,

24. glb = 0;

0≤

glb = 12 (−1 −

 1 3 11 1 10

√

5)

< 0 + 0.001

2n−1

 2k

25. Let  > 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 + . Suppose on the contrary that there is no such number in S. We then have m +  ≤ x for all x ∈ S. This makes m +  a lower bound for S. But this cannot be, for then m +  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

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SECTION 11.1 Since b ∈ S,

27. Let c = lub S.

b ≤ c. Since b is an upper bound for S,

c ≤ b.

589

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 1 − 2 < x20 − 2. n n Then,

2  2x0 2x0 1 1 − 2 < x20 − 2 =⇒ 2 < x20 − + 2 = x0 − n1 n n n n

(c) If x20 < 2, then choose a positive integer n such that 2x0 1 + 2 < 2 − x20 . n n Then x20 +

 2 2x0 1 + 2 < 2 =⇒ x0 + n1 < 2 n n

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

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SECTION 11.2

36.

a2

a3

a4

a5

a6

9 4

5 3

6 5

1 2

−3

(b) a20 = 94 , (c) lub S = 4,

a30 = −3,

a40 = 65 ,

a7

a8

a9

a10

4

9 4

5 3

6 5

a50 =

9 4

glb S = −3

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

37. (a)

(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 , n = 1, 2, 3, . . . 2n − 1

4.

an =

5.

an =

n2 + 1 , n = 1, 2, 3, . . . n

6.

an = (−1)n

 7. an =

n 1/n,

if n = 2k − 1 if n = 2k,

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

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

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SECTION 11.2  8. an =

n

if n = 2k 2

1/n ,

9. decreasing;

591

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

if n = 2k − 1,

bounded below by 0 and above by 2

10. not monotonic; bounded below by −1 and above by 12 . 11.

n + (−1)n 1 = 1 + (−1)n : not monotonic; n n

bounded below by 0 and above by

3 2

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 n2 1 1 =n−1+ : increasing; bounded below by but not bounded above n+1 n+1 2 √ √ 16. increasing; bounded below by 2, but not bounded above: n2 + 1 > n 15.

17.

18.

4n 1 2 and decreases to 0: increasing; = 4n2 4n2 + 1 1 + 1/4n2 √ bounded below by 45 5 and above by 2 √

2n 2n 1 = n 2 = n +1 (2 ) + 1 2 +

4n

19. increasing; 20.

√

1 2n

decreasing; bounded below by 0 and above by 25 .

.

bounded below by

n2 1 = 1 n3 + 1 n +

1 n4

and

2 51

but not bounded above

1 1 + decreases to 0 =⇒ n n4

√ increasing; bounded below by

2 , 2

but not bounded above. 2n 2 =2− increases toward 2: increasing; bounded below by 0 and above by ln 2. n+1 n+1 √ 2 2 22. increasing (n ≥ 2); bounded below by 10 , but not bounded above. 3 21.

23. decreasing; bounded below by 1 and above by 4 24. not monotonic; not bounded below and not bounded above 25. increasing; bounded below by 26. decreasing (since

√

3 and above by 2

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

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SECTION 11.2

√ √ 27. (−1)2n+1 n = − n: decreasing; √ 28.

n+1 √ = n

1+

bounded above by −1 but not bounded below

√ 1 decreasing; bounded below by 1 and above by 2 n

29.

2n − 1 1 = 1 − n : increasing; n 2 2

30.

1 1 3 − = decreasing; bounded below by 0 and above by 2n 2n + 3 2n(2n + 3)

bounded below by

31. consider sin x as x → 0+ : decreasing;

1 and above by 1 2

bounded below by 0 and above by 1

32. not monotonic; bounded below by − 12 and above by 33. decreasing;

3 10 .

bounded below by 0 and above by

1 4

5 6

x 4 is an increasing function on [4, ∞).); bounded below by ln x ln 4 but not bounded above.

34. increasing (because

35.

1 1 1 − = : decreasing; n n+1 n (n + 1)

bounded below by 0 and above by

1 2

36. not monotonic; bounded below by −1 and above by 1 37. Set f (x) =

ln x . x

Then, f  (x) =

bounded below by 0 and above by

1 − ln x e: decreasing;

ln 3.

38. not monotonic; not bounded below nor above (because exponentials grow faster than polynomials). 39. Set an =

3n . (n + 1)2

bounded below by 40.

1 − ( 12 )n = 2n − 1 ( 12 )n

Then, 3 4

an+1 =3 an



n+1 n+2

2 > 1: increasing;

but not bounded above. increasing; bounded below by 1 but not bounded above.

41. For n ≥ 5 an+1 n! 5n+1 5 · an . Since a2 = 1 + √ Assume that ak = 1 + ak−1 > ak−1 . Then ak+1 = 1 +

√

ak > 1 +

√

√

a1 = 2 > 1, 1 ∈ S.

ak−1 = ak .

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

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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

and solving for

√

an

√ 2 √ ( an ) − an − 1 < 0 √ √ it follows that an < 12 (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:

6.

converges to 1:

7. converges to 0: 8. converges to 0: 9. converges to 0:

10.

diverges:

12. diverges:

n−1 1 =1− →1 n n

n+1 1 1 = + 2 →0 2 n n n π  π → 0, so sin → sin 0 = 0 2n 2n 0<

2n 2n 1 < = n →0 n n 4 +1 4 2

n2 n2 n ≥ = n+1 2n 2 √

4n 4n →∞ ≈ n n2 + 1

13. converges to 0 14. converges to 0:

4n 4n n < n = n−2 → 0 2n + 106 2 2

11.

diverges

n + (−1)n (−1)n =1+ →1 n n

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SECTION 11.3

15. converges to 1:

nπ π → 4n + 1 4

16. converges to 0:

√ 1010 n 1010 √ →0 =√ n+1 n + 1/ n

17. converges to

nπ π → tan = 1 4n + 1 4



2n → 2, n+1

1√ 2: 2

√

so ln

diverges:

cos nπ = (−1)

24. converges to

22.

25. converges to 0 :

ln n − ln (n + 1) = ln

26. converges to 1:

2n − 1 1 = 1 − n =⇒ 1 2n 2 √

1 converges to : 2

n+1 1 √ = 2 2 n

 29. converges to e2 :

31. diverges;

diverges:

n5 =n 17n4 + 12



1 17 +

 12 n4

4=2 

30. converges to

→ ln 2

√ 1 √ → 0 so e1/ n → e0 = 1 n

√

√



→1

6 n4

n

23. converges to 1:

27.

2n n+1

n2 1 1 →√ = 4 4 2 2n + 1 2 + 1/n

1 − n14 n4 − 1 = n4 + n − 6 1 + n13 −

20. converges to 1:

21.

tan

(2n + 1)2 4 + 4/n + 1/n2 4 = → (3n − 1)2 9 − 6/n + 1/n2 9

4 : 9

18. converges to ln 2 :

19. converges to

so

1+ 

e:

1 n

2n

1 1+ n

2 n > n3

since



=

1+

n n+1

1 1 → n 2

33. converges to 0:

converges to 0:

n 2

 1 1+ → e2 n 

n/2 =

1+

for n ≥ 10,

1 n

n →

√

e

2n n3 > 2 =n 2 n n

2 ln 3n − ln(n2 + 1) = ln

| sin n| 1 √ ≤√ n n

→ ln 1 = 0

28.

 32. converges to ln 9 :



9n2 2 n +1

 → ln 9

1 1 1 − = →0 n n+1 n(n + 1)

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SECTION 11.3 34.

597

n → 1, arctan 1 = π/4 n+1

converges to π/4:

  √ n2 + n + n n 1 2 2 35. converges to 1/2: =√ → n +n−n= n +n−n √ 2 2 2 n +n+n n +n+n √ 36. converges to 2:

√ 4n2 + n = 4 + (1/n) → 4 = 2. n

37. converges to π/2 38. converges to 31/9: let s = 3.444 · · ·. Then 10s − s = 31 and s = 31/9. 1−n → −1; arcsin (−1) = π/2. n

39. converges to −π/2:

(n + 1)(n + 4) n2 + 5n + 5 = 2 → 1. (n + 2)(n + 3) n + 5n + 6

40. converges to 1:

41. (a)

√ n

n→1

(b)

n 42. (a) √ →e n n! 43. b <

√ n

3n →0 n!

(b) does not converge

an + bn = b

n

(a/b)n + 1 < b

√ n

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

by the pinching theorem. 44. (a) −1 < r ≤ 1 45. Set  > 0.

(b) −1 < r < 1

Since an → L, there exists N1 such that n ≥ N1 ,

if

then

|an − L| < /2.

Since bn → M , there exists N2 such that n ≥ N2 , then

if

|bn − M | < /2.

Now set N = max {N1 , N2 }. Then, for n ≥ N , |(an + bn ) − (L + M )| ≤ |an − L| + |bn − M | < 46. Let  > 0, choose k such that n ≥ k =⇒ |an − L| < Then for n ≥ k, Therefore  47. Since

  + = . 2 2

 . |α| + 1

|αan − αL| = |α||an − L| ≤ (|α| + 1)|an − L| < 

α an → α L. 1 1+ n



 →1

n

and

1 1+ n



n+1

1 1+ n

→ e,  =

1 1+ n

n 

1 1+ n

 → (e)(1) = e.

√ n

an + bn → b

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SECTION 11.3

48. (a)

(b)

(c)

If k = j,

If k < j,

If k > j,

1 1 + · · · + α0 · k αk n n an = → 1 1 βk βk + βk−1 · + · · · + β0 · k n n αk + αk−1 ·

an =

an =

αk + αk−1 ·

1 1 + · · · + α0 · k n n

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

1 nk

αk · nk−j + αk−1 · nk−1−j + · · · + α0 ·

1 nj

βj + βj−1 ·

1 1 + · · · + β0 · j n n

→

0

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  > 0. By Theorem 11.1.4 there exists ak such that ak < L + . Since the sequence is non-increasing, an ≤ ak for all n ≥ k. Thus, L ≤ an < L +  or |an − L| <  and

for all

n≥k

an → L. If an → L,

50. Let  > 0.

then there exists a positive integer k such that |an − L| <  for all n ≥ k

If n ≥ k,

then 2n ≥ k and 2n − 1 ≥ k,

and thus

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

and if m ≥ k2 , Let k = max{2k1 , 2k2 − 1}.

then |om − L| = |a2m−1 − L| < 

If n ≥ k then

either an = a2m with m > k1 or an = a2m−1 with m ≥ k2 In either case, 51. Let  > 0.

|an − L| < .

This shows that an → L.

Choose k so that, for n ≥ k, L −  < an < L + ,

L −  < cn < L +  and an ≤ bn ≤ cn .

For such n, L −  < bn < L + . 52. Let M be a bound for {bn }.

Then |an bn | ≤ |an |M.

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

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SECTION 11.3

599

53. Let  > 0. Since an → L, there exists a positive integer N such that L −  < an < L +  for all n ≥ N. Now an ≤ M for all n, so L −  < M, or L < M + . Since  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  > 0. There exists a positive integer N such that |an − 0| <  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  > 0. There exists a positive integer N1 such that |an − L| <  for all n > 2N1 − 1, and there exists a positive integer N2 such that |bn − L| <  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| <  =⇒ cn → L.  57. By the continuity of f, f (L) = f 58.

 lim an = lim f (an ) = lim an+1 = L.

n→∞

n→∞

n→∞

2n 2 2 2 2 2 4 = · · · · · = 2 · · (terms that are ≤ 1) ≤ . n! 1 2 3 n n n 4 →0 n

Since

and

0<

2n 4 ≤ , n! n

2n →0 n!

as well.

1 59. Set f (x) = x1/p . Since → 0 and f is continuous at 0, it follows by Theorem 11.3.12 that n  1/p 1 → 0. n 60. Since |an − L| = |(an − L) − 0| = ||an − L| − 0|, |an − L| <  iff

|(an − L) − 0| <  iff

So an → L iff

an − L → 0

61.

an = e1−n → 0

63.

an =

65.

an =

67.

L = 0,

69.

L = 0,

iff

||an − L| − 0| < ,

|an − L| → 0. 62.

diverges

1 →0 n!

64.

an = 1 ·

1 [1 − (−1)n ] diverges 2

66.

an =

n = 32

68.

1 √ → 0. n

1 √ < 0.001 for n ≥ 10002 + 1 n

n=4

70.

n10 → 0. 10n

n10 < 0.001 for n ≥ 15 10n

1 2 n−1 1 · ··· = 2 3 n n

converges to 0

2n − 1 →2 2n−1

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SECTION 11.3

71.

L = 0,

n=7

72.

2n → 0. n!

2n < 0.001 for n ≥ 10 n!

73.

L = 0,

n = 65

74.

ln n → 0. n

ln n < 0.001 for n ≥ 9119 n

√ 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 = 12 (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,

(b)

True for n = 1.

(c)

an+1 2 − an 2 = 6 + an − an 2 = (3 − an )(2 + an ) ≥ 0 Since ak ≥ 0

Assume true for n.

a5 ∼ = 2.9983, a6 ∼ = 2.9997 √ √ Then an+1 = 6 + an−1 ≤ 6 + 3 = 3 since 0 ≤ an ≤ 3.

this implies an+1 ≥ an

for all k,

(d) an → 3 77. (a)

a2

a3

a4

a5

a6

0.5403

0.8576

0.6543

0.7935

0.7014

a7

a8

0.7640 0.7221 ∼ 0.739085. (b) L is a fixed point of f (x) = cos x, that is, cos L = L; L = a2 ∼ = 1.5403, a3 ∼ = 1.5708, a4 ∼ = a5 ∼ = ··· ∼ = a10 ∼ = 1.5708. (b) L ∼ = 1.570796 Let f (x) = x + cos x. L must satisfy L = f (L), π and cos L = 0. Indeed, the L we found is just ∼ = 1.570796327 2

a9

a10

0.7504

0.7314

78. (a)

so L = L + cos L,

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 3 (b) L = L+ which implies L2 = 3 or L = 3. 2 L

2. The Newton-Raphson method applied to the function f (x) = x2 − R an = an−1 −

= 3 & 4. (a) f (x) = x3 − 8, (c) f (x) = ln x − 1,

so

gives

a2 − R f (an−1 ) = an−1 − n−1  f (an−1 ) 2an−1

1 1 R 1 an−1 + = 2 2 an−1 2 so

1.732051

xn → 2 xn → e

 an−1 +

R an−1

 ,

n = 2, 3, . . . .

(b) f (x) = sin x − 12 ,

so

xn →

π 6

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SECTION 11.4

601

SECTION 11.4 1.

22/n = (21/n )2 → 12 = 1

converges to 1:

3. converges to 0:

4. converges to 0:

5. converges to 0:

6.

converges to 0:

8.

converges to 1:

3n = 4n

 n 3 →0 4

n/(n+2)  n1/(n+2) = n1/n →1 n+1 n

11. converges to 0:



0

−n

: 

n



3n 4n

 n 3 = 12 → 12(0) = 0 4

1 e−2n 1 − → 2 2 2

e2x dx =

1 1− n



n =

0



n

integral = 2 0



n

18. converges to 0:

(−1) 1+ n

n

1 →1 en

→ e−1

16.

(by (11.4.7)

diverges.

π dx −1 = 2 tan n → 2 =π 1 + x2 2

1 e−n + →0 dx = − n n 2

−nx

e 0

21. converges to 0:



e−x dx = 1 −

converges to 1:

converges to 1:

nα/n = (n1/n )α → 1α = 1

(n + 2)1/n → e0 = 1. 

19.

converges to 1:

1

it follows that

17. converges to π:

9.

x100n (x100 )n = →0 n! n!

(n + 2)1/n = e n ln(n+2) and, since

  1 ln (n + 2) n+2 ln (n + 2) = → (0)(1) = 0, n n+2 n

13. converges to 1:

15.

converges to 0:

→ ln(1) = 0

3n+1 = 12 4n−1

1 : 2

7.



ln

14. converges to e

e−α/n → e0 = 1

log10 n 1 ln n = · →0 n ln 10 n

  ln (n + 1) ln (n + 1) n+1 = → (0)(1) = 0 n n+1 n

10. converges to 0:

−1

converges to 1:

 n  n 2 2 0< < →0 n 3

for n > 3,



12. converges to

2.

recall (11.4.6) ln (n2 ) ln n =2 → 2(0) = 0 n n

20.

converges to 0:

n2 sin nπ = 0

for all n

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SECTION 11.4 

22. converges to π :

23. diverges:

24.

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1−1/n

−1+1/n

Since

lim

x→0

    dx 1 1 −1 −1 √ = sin − sin → sin−1 (1) − sin−1 (−1) = π 1− −1 + n n 1 − x2

sin x = 1, x

n π sin (π/n) sin = → 1. Therefore, π n π/n n π π n2 sin = nπ sin → nπ. n π n

diverges

25.



5 16

n →0

x 3n  x n 3 = 1+ → (ex )3 = e3x n n



26. converges to e3x :

converges to 0:

5n+1 = 20 42n−1

1+

n+2 (−1)  n  n 1+ n+1 1 e−1 n+2 = 1− =  = e−1 2 → n+2 n+2 1 (−1) 1+ n+2 

27. converges to e−1 :



1

dx 2 √ = 2 − √ → 2. x n

28. converges to 2: 1/n

 29. converges to 0:

n+1

0<

e−x dx ≤ e−n [(n + 1) − n] = e−n → 0 2

2

2

n





n

30. converges to 1:

1 1+ 2 n

31. converges to 0:

 n n nn →0 2 = n 2n 2 

x

cos e dx 0

33. converges to ex :  34. diverges:

1 1+ n

35. converges to 0:  t+

37.

converges:

→ (e1 )0 = 1

since

→

0

n →0 2n

cos ex dx = 0

0

use (11.4.7) n2



  n n n 1 1 n = 1+ >2 , 1+ ≈e>2 n n

    1/n  1/n  1/n 2   2 2 sin x dx ≤ | sin x | dx ≤ 1 dx = → 0   −1/n  n −1/n −1/n

 n x n x/t = tn 1 + ; n n

36.

n2 1/n



1/n

32. converges to 0:

=

1 1+ 2 n

sin(6/n) →2 sin(3/n)

converges to 0 if t < 1,

38.

converges to ex if t = 1,

converges:

arctan n →0 n

diverges if t > 1.

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SECTION 11.4 39.

√

n+1−

√

603

√ √ √  n + 1 − n √ 1 n= √ n+1+ n = √ √ √ →0 n+1+ n n+1+ n

√ n2 + n − n 2 n 1 1 2 40. n +n−n= √ ( n + n + n) = √ = → 2 2 2 n +n+n n +n+n 1 + 1 + 1/n 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. d < (cn + dn )1/n < (2dn )1/n = 21/n d → d,

42. Since 0 < c < d,

so by the pinching theorem

(cn + dn )1/n → d. 43. By the hint,

44. diverges:

lim

n→∞

1 1 + 2 + ... + n n(n + 1) 1 + 1/n = . = lim = lim n→∞ n→∞ n2 2n2 2 2

12 + 22 + · · · + n2 n(n + 1)(2n + 1) 2n3 + 3n2 + n = = 2 →∞ (1 + n)(2 + n) 6(1 + n)(2 + n) 6n + 18n + 12

45. By the hint,

13 + 23 + . . . + n3 n2 (n + 1)2 1 + 2/n + 1/n2 1 = . = lim = lim n→∞ n→∞ 4(2n4 + n − 1) n→∞ 8 + 4/n3 − 4/n4 2n4 + n − 1 8 lim

46. Here we show that every convergent sequence is a Cauchy sequence. Let  > 0. If an → L, then there exists a positive integer k such that |ap − L| < With m, n ≥ k

 2

for all p ≥ k

we have |am − an | ≤ |am − L| + |L − an | = |am − L| + |an − L| < mn+1 − mn =

47. (a)

  + = . 2 2

1 1 (a1 + · · · + an + an+1 ) − (a1 + · · · + an ) n+1 n n

     1 = nan+1 − (a1 + · · · + an ) n(n + 1) >0

since {an } is increasing.

(b) We begin with the hint |a1 + · · · + aj |  mn < + n 2



Since j is fixed, |a1 + · · · + aj | →0 n

n−j n

 .

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December 2, 2006

SECTION 11.4 and therefore for n sufficiently large |a1 + · · · + aj |  < . n 2 Since  2



n−j n

 <

 , 2

we see that, for n sufficiently large, |mn | < . This shows that mn → 0. 48. (a) Since {an } converges, it is a Cauchy sequence (see Exercise 46), so given  > 0 we can find k such that |an − am | <  for n, m ≥ k. In particular,

|an+1 − an | < ,

so

lim (an − an−1 ) = 0.

n→∞

(b) {an } does not necessarily converge. For example let an = ln n. This diverges, but   n an − an−1 = ln n − ln(n − 1) = ln → ln 1 = 0 n−1 49. (a) Let S be the set of positive integers n (n ≥ 2) for which the inequalities hold. Since √ 2 √ √ 2 √ √ 2 b − 2 ab + a = b − a > 0, it follows that

a+b √ > ab and so a1 > b1 . Now, 2 a1 + b1 a2 = < a1 and b2 = a1 b1 > b1 . 2

Also, by the argument above, a2 =

a1 + b1 > a1 b1 = b2 , 2

and so a1 > a2 > b2 > b1 . Thus 2 ∈ S. Assume that k ∈ S. Then ak + bk ak + ak ak+1 = < = ak , bk+1 = ak bk > b2k = bk , 2 2 and ak+1 =

ak + bk > ak bk = bk+1 . 2

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 = lim an , n→∞

Lb = lim bn . Then n→∞

an−1 + bn−1 an = 2 50.

implies

La =

La + Lb 2

and La = Lb .

    1 100 5 100 e − 1 + 100 e5 − 1 + 100 ∼ ∼ = 0.004995 : within 0.01%; = 0.11395 : within 12% e e5     1000 1000 1 5 e5 − 1 + 1000 e − 1 + 1000 ∼ ∼ = 0.0004995 : within 0.05%; = 0.01238 : within 1.3% e e5

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SECTION 11.5

605

51. The numerical work suggests L ∼ = 1. Justification: Set f (x) = sin x − x2 . Note that f (0) = 0 and f  (x) = cos x − 2x > 0 for x close to 0. Therefore sin x − x2 > 0 for x close to 0 and sin 1/n − 1/n2 > 0 for n large. Thus, for n large, 1 1 1 (| sin x| ≤ |x| for all x) < sin < n2 n n  1/n  1/n  1/n 1 1 1 < sin < n2 n n 1/n  2  1 1 1 < sin < 1/n . 1/n n n 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. (1 + 1/n)1/k − 1 ; 1/n

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

(1 + 1/n)1/k − 1 (1 + h)1/k − 1 = lim = f  (1), n→∞ h→0 1/n h lim

where f (x) = x1/k . 53. (a) (b)

Since f  (x) = (1/k)x(1/k)−1 ,

a3

a4

a5

a6

a7

a8

a9

a10

2

3

5

8

13

21

34

55

r1

r2

r3

r4

r5

r6

1

2

1.5

1.667

1.600

1.625

f  (1) = 1/k.

(c) Following the hint, 1+

1

1 an−1 an + an−1 an+1 = 1 + an = 1 + = = = rn . an an an an−1

rn−1

Now, if rn → L, then rn−1 → L and

√ 1 1+ 5 ∼ 1 + = L which, since L > 1, implies L = = 1.618034. L 2

54. With the partition {0, n1 , n2 , . . . , nn } and f (x) = x, we have  

    n  n   1 1 2 n 1 1 2 an = f (xi )Δxi , + + ··· + = f +f + ··· + f = n n n n n n n n i=1 



1

so it is a Riemann sum for

x dx,

and therefore lim an = n→∞

0

1

x dx = 0

1 2

SECTION 11.5 (We’ll use  to indicate differentiation of numerator and denominator.) 1.

lim

x→0+

√ sin x  √ = lim 2 x cos x = 0 x x→0+

2.

lim

x→1

ln x  1/x = −1 = lim 1 − x x→1 −1

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606 3.

5.

7.

SECTION 11.5

11.

13.

√

ex − 1  = lim (1 + x)ex = 1 x→0 ln (1 + x) x→0

4.

cos x  − sin x 1 = = lim sin 2x x→π/2 2 cos 2x 2

6.

lim

lim

x→π/2

2x − 1  = lim 2x ln 2 = ln 2 x→0 x→0 x lim

x1/2 − x1/4  9. lim = lim x→1 x→1 x−1 10.

December 2, 2006



1 −1/2 1 −3/4 − x x 2 4

8.  =

x→4

lim

x→a

1

√ x−2  1 2 x = lim = x→4 1 x−4 4

lim

x−a  1 1 = lim = xn − an x→a nxn−1 nan−1

tan−1 x  = lim x→0 x→0 x lim

1 1+x2

1

=1

1 4

ex − 1  ex = lim =1 x→0 x(1 + x) x→0 1 + 2x lim

ex − e−x  ex + e−x = lim =2 x→0 x→0 cos x sin x lim

lim

x→0

12.

1 − cos x  sin x = lim =0 x→0 x→0 3 3x lim

x + sin πx  1 + π cos πx 1+π = lim = x→0 x − sin πx 1 − π cos πx 1−π 



14.

ax − (a + 1)x  ax ln a − (a + 1)x ln(a + 1) = lim = ln x→0 x→0 x 1

15.

ex + e−x − 2  ex − e−x  ex + e−x 1 = lim = lim = x→0 1 − cos 2x x→0 2 sin 2x x→0 4 cos 2x 2

16.

1 − x+1  x − ln(x + 1)  1 (x+1)2 = lim = lim = x→0 1 − cos 2x x→0 2 sin 2x x→0 4 cos 2x 4

17.

tan πx  π sec2 πx = lim =π x x→0 e − 1 x→0 ex

18.

cos x − 1 + x2 /2  − sin x + x  − cos x + 1  sin x 1 = lim = lim = lim = x→0 x→0 x→0 x→0 24x x4 4x3 12x2 24

19.

1 + x − ex  1 − ex −ex 1  = lim = lim =− x x x x x→0 x(e − 1) x→0 xe + e − 1 x→0 xe + 2ex 2

20.

ln(sec x)  tan x  sec2 x 1 = lim = lim = 2 x→0 x→0 2x x→0 x 2 2

21.

x − tan x  1 − sec2 x  −2 sec2 x tan x −2 sec2 x = lim = lim = lim = −2 x→0 x − sin x x→0 1 − cos x x→0 x→0 sin x cos x

22.

xenx − x  enx + nxenx − 1  enx (2n + n2 x) 2 = lim = lim = x→0 1 − cos nx x→0 x→0 n sin nx n2 cos nx n

23.



√ 1 − x2 1 − x2 2 1√ lim− √ = lim− = 6 = 3 x→1 x→1 1−x 3 3 1 − x3

lim

a a+1

lim

1

1

lim

lim

lim

lim

lim

lim

lim

since

lim−

x→1

1 − x2  2x 2 = lim = 1 − x3 x→1− 3x2 3

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SECTION 11.5 24.

25.

lim

x→0

2x − sin πx =0 4x2 − 1 ln (sin x)  − cot x  csc2 x 1 = lim = lim =− (π − 2x)2 x→π/2 4(π − 2x) x→π/2 −8 8

lim

x→π/2

√

26.

27.

lim √

x→0+

lim

x→0

lim

x→0

29.

1 √ 2 x √ cos√ x 1 √ + 2 x 2 x

= lim

x→0+

1 1 √ = 2 1 + cos x

cos x − cos 3x  − sin x + 3 sin 3x  − cos x + 9 cos 3x = lim = lim =4 x→0 x→0 2 cos(x2 ) − 4x2 sin(x2 ) sin(x2 ) 2x cos(x2 ) √

28.

x  √ = lim x + sin x x→0+

a+x− x

√

a−x



= lim

√1 2 a+x

+

√1 2 a−x

1

x→0

√ a 1 =√ = a a

sec2 x − 2 tan x  2 sec2 x tan x − 2 sec2 x = lim 1 + cos 4x −4 sin 4x x→π/4

lim

x→π/4



= lim

x→π/4

2 sec4 x + 4 sec2 x tan2 x − 4 sec2 x tan x 1 = −16 cos 4x 2

−x 1 1 − √1−x x − arcsin x  2  (1−x2 )3/2 30. lim = lim = lim x→0 x→0 3 sin2 x cos x x→0 6 sin x cos2 x − 3 sin3 x sin3 x



= lim

x→0

31.

32.

lim

x→0

6 cos3

x − 21 sin x cos x

sin−1 x  = lim x→0 x→0 x lim

x→∞

1 6

√ 1 1−x2

1

=1

π/2 − tan−1 x  x2 = lim =1 x→∞ 1 + x2 1/x

ln(1 − n1 ) ln(1 − x)  1 = −1 = lim = lim n→∞ sin( 1 ) sin x x→0+ x→0+ (1 − x) cos x n

34. −1 :

35. 1 :

=−

tan−1 x  1 1 +˙4x2 1 = lim = −1 2 x→0 tan 2x 1+x 2 2

lim

33. 1 :

36.

−1−2x2 (1−x2 )5/2 2

lim

lim

1 1/x t  = lim = lim = lim (1 + t) = 1 x[ ln (x + 1) − ln x ] x→∞ ln (1 + 1/x) t→0+ ln (1 + t) t→0+

lim

sinh(π/n) − sin(π/n) sinh(π/x) − sin(π/x)  sinh u − sin u = lim = lim x→∞ u→0+ sin3 (π/n) sin3 (π/x) sin3 u

x→∞

1 : 3 

n→∞

= lim+ u→0

2 1 cosh u − cos u  sinh u + sin u cosh u + cos u  = = = lim+ = lim+ 3 2 6 3 u→0 6 sin u cos2 u − 3 sin u u→0 6 cos3 u − 21 sin u cos u 3 sin2 u cos u

607

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x3 =0 x→0 4x − 1

39.

lim

43.

December 2, 2006

lim

x→0 π 2

lim

x→0

38.

x =1 − arccos x

tanh x =1 x

lim (2 + x + sin x) = 0,

x→0

 44.

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SECTION 11.5

37.

41.

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1/n

lim n a

n→∞

lim

x→0

4x does not exist sin2 x

40.

√ 2x − 2 lim √ =0 x→2+ x−2

42.

1 + cos 2x =4 x→π/2 1 − sin x lim

lim (x3 + x − cos x) = 0

x→0

  a1/x ln a − x12 a1/x − 1   1 = ln a − 1 = lim = lim x→∞ x→∞ 1/x − x2 

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

x→0

cos ax − 1  −a sin ax  −a2 cos ax a2 = lim = lim =− 2 x→0 x→0 2x 4x 4 4

Now, −

46.

a2 = − 4 implies 4

a = ±4.

sin 2x + ax + bx3  2 cos 2x + a + 3bx2 = lim 3 x→0 x→0 x 3x2 lim

−4 sin 2x + 6bx  −8 cos 2x + 6b = lim =0 x→0 x→0 6x 6 4 =⇒ a = −2, b = 3 

= lim

need a = −2 to keep numerator 0 if 6b = 8

47. Recall lim (1 + x)1/x = e: + x→0

  x − (1 + x) ln (1 + x)  (1 + x)1/x − e  1/x lim = lim (1 + x) x x2 + x3 x→0+ x→0+ = e lim

x→0+



= e lim

x→0+



= e lim

x→0+

x − (1 + x) ln (1 + x) x2 + x3 − ln (1 + x) 2x + 3x2 e −1/(1 + x) =− 2 + 6x 2

f (x + h) − f (x − h)  f  (x + h) − f  (x − h)(−1) = lim = f  (x) h→0 h→0 2h 2

48. (a) lim

(note that here we differentiated with respect to h, not x. ) f (x + h) − 2f (x) + f (x − h)  f  (x + h) − f  (x − h) = lim 2 h→0 h→0 h 2h

(b) lim 

f  (x + h) + f  (x − h) = f  (x) h→0 2

= lim

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SECTION 11.5 49.

1 x→0 x



x

lim



f (t) dt = lim

x→0

0

50. (a) lim

x→0

f (x) = f (0) 1

Si(x)  sin x = lim =1 x→0 x x

(b)

lim

x→0

Si(x) − x  sin x/x − 1 sin x − x = lim = lim x→0 x→0 x3 3x2 3x3 

= lim

x→0



= lim

x→0

cos x − 1 9x2 − sin x 1 =− 18x 18

C(x)  cos2 x = lim =1 x→0 x→0 x 1 C(x) − x  cos2 x − 1  −2 cos x sin x  −2 cos2 x + 2 sin2 x 1 (b) lim = lim = lim = lim =− 3 2 x→0 x→0 x→0 x→0 x 3x 6x 6 3  x f (t) dt f (x)  52. (a) lim a =0 = lim  x→a x→a f (x) f (x)  x f (t) dt f (x)  f (k−1) (x)   a (b) Similarly, lim = lim  = · · · = lim (k) =0 x→a x→a f (x) x→a f f (x) (x)

51. (a) lim



√

b

x3 (b − x2 ) dx = 2 bx − 53. A(b) = 2 3 0 √ T (b) b b 3 Thus, lim = 4 √ = . b→0 A(b) 4 b b 3 54. T (θ) =

1 (1 − cos θ) sin θ; 2 lim

θ→0+

S(θ) =

√b = 0

4 √ b b 3

and

T (b) =

√ 1 √  2 b b = b b. 2

θ 1 − sin θ: 2 2

T (θ) (1 − cos θ) sin θ  (1 − cos θ) cos θ + sin2 θ cos θ − cos 2θ = lim = lim = lim + + + S(θ) θ − sin θ 1 − cos θ 1 − cos θ θ→0 θ→0 θ→0 

=

− sin θ + 2 sin 2θ − sin θ + 4 sin θ cos θ = lim+ =3 sin θ sin θ θ→0 y

55. (a)

f (x) → ∞ as x → ±∞

15 f 10 5 -5

5

x

(b) f (x) → 10 as x → 4 Confirmation: lim √ x→4

x2 − 16 2x  = lim = lim 2 x2 + 9 = 10 −1/2 x→4 x2 + 9 − 5 x→4 x (x2 + 9)

609

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SECTION 11.6 y

56. (a)

lim f (x) = 0

x→±∞

0.17

−10

(b) f (x) →

10

1 as x → 0; 6

lim

x→0

x

x − sin x  1 − cos x  sin x 1 = lim = lim = x→0 x→0 6x x3 3x2 6

y

57. (a)

f (x) → 0.7 as x → 0

0.8 f

0.6

-2

-1

1

(b) Confirmation: lim

x→0

2

x

2sin x − 1  ln(2) 2sin x cos x = lim = ln 2 ∼ = 0.6931 x→0 x 1 g(x) → −1.6 as x → 0

y

58. (a)

x

−1

−2

(b) Confirmation: lim

x→0

3cos x − 3  3cos x (− sin x) ln 3 sin x = lim = − lim 3cos x ln 3 x→0 x→0 x2 2x 2x =−

3 ln 3 ∼ = −1.6479 2

SECTION 11.6 (We’ll use  to indicate differentiation of numerator and denominator.) 1.

3.

lim

x→−∞

lim

x→∞

x2 + 1  2x = lim =∞ x→−∞ 1−x −1

x3 1 = −1 = lim x→∞ 1/x3 − 1 1 − x3

1 5. lim x sin = lim x→∞ x h→0+ 2



   sin h 1 =∞ h h

2.

4.

lim

x→∞

20x =0 +1

x2

x3 − 1 = −∞ x→∞ 2 − x lim

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SECTION 11.6 ln(xk )  k/x = lim =0 x→∞ x→∞ 1 x

   sin 5x  cos x  tan 5x 1 7. lim = lim = − π− tan x sin x cos 5x 5 x→ π x→ 2 2

6.

lim

lim π−

x→ 2

8. 9. 10.

11.

12.

lim π−

x→ 2

cos x  sin x 1 = = lim cos 5x x→ π2 − 5 sin 5x 5

cos x  x  ln | sin x|  = lim sin1x = lim − (x cos x) = 0 x→0 x→0 − 2 x→0 1/x sin x x

x→0

lim x2x = lim (xx )2 = 12 = 1

x→0+

x→0+

[see (11.6.4)]

π sin πt  π cos πt = lim = lim =π + + x t→0 t 1 t→0

lim x sin

x→∞

(ln |x|)2  2 ln |x|  2 = lim 2x = 0 = lim = lim x→0 x→0 x→0 1/x −1/x 1/x x→0

lim x( ln |x| )2 = lim

x→0

ln x  1/x sin2 x sin x = lim = lim+ − = lim+ − · sin x = 0 2 cot x x→0+ − csc x x→0 x x x→0

lim

x→0+

√ lim

x→∞



x

0

2

ex =∞ e dt = lim x→∞ 1 

t2

1 + x2 = lim x→∞ x



1 +1=1 x2

1 1 lim − 2 x→0 sin2 x x

15.

 = lim

x2 − sin2 x  2x − 2 sin x cos x = lim x→0 2x2 sin x cos x + 2x sin2 x x2 sin2 x

= lim

2x − sin 2x 2 − 2 cos 2x  = lim x2 sin 2x + 2x sin2 x x→0 2x2 cos 2x + 4x sin 2x + 2 sin2 x

x→0

x→0



= lim

x→0 −4x2



= lim

x→0

16. Since 17.

and

lim (x ln | sin x|) = lim

1 13. lim x→∞ x 14.

sin 5x =1 sin x

since

−8x2

4 sin 2x sin 2x + 12x cos 2x + 6 sin 2x 8 cos 2x 1 = cos 2x − 32x sin 2x + 24 cos 2x 3

lim ln(| sin x|x ) = lim (x ln | sin x|) = 0

x→0

x→0

lim x1/(x−1) = e

x→1

since

sin x

lim ln x

x→0+



x→0

  ln x  1 lim ln x1/(x−1) = lim = lim = 1 x→1 x→1 x − 1 x→1 x

18. Take log: 

by Exercise 8, lim | sin x|x = e0 = 1

 = lim (sin x ln x) = lim + + x→0



= lim

x→0+

x→0

ln x csc x



1/x − sin2 x = lim = 0, − csc x cot x x→0+ x cos x

so

lim xsin x = e0 = 1

x→0+

611

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SECTION 11.6  cos

lim

19.

x→∞

1 x

  1

 x  ln cos 1 x = lim lim ln cos x→∞ x→∞ x (1/x)   sin (1/x)  = lim − =0 x→∞ cos (1/x)

x =1

since

20. Take log: lim ln (| sec x|cos x ) = lim cos x ln | sec x| = lim

x→π/2

x→π/2



x→π/2

ln | sec x| sec x

tan x = lim cos x = 0, x→π/2 sec x tan x x→π/2

= lim

lim

21.

x→0



x→0

so

2

ln(x2 + a2 )  = lim x→∞ x→∞ x2

lim ln(x2 + a2 )(1/x) = lim

x→∞

2x x2 +a2

2x

2

lim

x→0

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

= lim

x→0

 lim ln

x→∞

lim

x→∞



x2 − 1 x2 + 1

3

 = 3 lim ln x→∞

x→∞

27.

x2 − 1 x2 + 1

sin x + x cos x =0 2 cos x − x sin x

 =0

      √x2 + 2x + x √ x2 + 2x − x = lim x2 + 2x − x x→∞ x2 + 2x + x = lim √

26.

= 0,

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

25.

1 1 = 1 + 1 + ln (1 + x) 2

x→∞

23.

24.

lim | sec x|cos x = e0 = 1

x→π/2

 1 x − ln (1 + x)  x 1 = lim − = lim x→0 x ln (1 + x) x→0 x + (1 + x) ln (1 + x) ln (1 + x) x = lim

22. Take log:

so

x2

2x 2 = lim =1 x→∞ + 2x + x 1 + 2/x + 1

  a bx a x b lim 1 + = lim 1 + = (ea )b = eab . x→∞ x→∞ x x  1/ ln x lim x3 + 1 = e3

x→∞

since 

  1/ ln x  ln x3 + 1  3 lim ln x + 1 = lim = lim x→∞ x→∞ x→∞ ln x 28. Take log: so

ln(ex + 1)  lim = lim x→∞ x→∞ x x

1/x

lim (e + 1)

x→∞

=e

ex ex +1 

1

ex = 1, x→∞ ex

= lim



3x2 x3 + 1 1/x

 = lim

x→∞

3 = 3. 1 + 1/x3

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SECTION 11.6 29.

lim (cosh x)1/x = e since

x→∞

lim ln [(cosh x)1/x ] = lim

x→∞

x→∞



1 30. Take log: lim 3x ln 1 + x→∞ x 3x  1 so lim 1 + = e3 x→∞ x  31.

613

lim

x→0

1 1 − sin x x

 = lim

x→0



= lim

x→0

 = 3 lim

ln (cosh x)  sinh x = lim = 1. x→∞ x cosh x

  ln 1 + x1 1 x

x→∞

−1/x2 1+1/x = 3 lim x→∞ − 12 x 

= 3,

x − sin x  1 − cos x = lim x→0 x sin x sin x + x cos x sin x =0 2 cos x − x sin x ex +3

32. Take log: 33.

ln(ex + 3x)  x = lim e +3x = 4, so lim (ex + 3x)1/x = e4 x→0 x→0 x→0 x 1   1 x − 1 − x ln x  − ln x x lim = lim − = lim x→1 x→1 (x − 1) ln x x→1 (x − 1)(1/x) + ln x ln x x − 1 lim

= lim

x→1

34.

35.

√

2:

take log:

36.

0:

 lim

x→0

1 + 2x 2

1/x

1

= e2

ln 2

=

√

2

nk →0 n→∞ 2n lim

  1 ln (ln n)1/n = ln (ln n) → 0 n

37. 1 : 38. 0:

2x ln 2 x ln 2 ln(1 + 2x ) − ln 2  ; lim = lim 1 + 2 = x→0 x→0 x 1 2

1 1 ln n ln = − →0 n n n

0:

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

lim

n→∞

ln n  1/n 1 = lim = lim =0 n→∞ p np−1 n→∞ p np np



1/n 

1 ln n ln (n + 1) ln [n(n + 1)] = + →0 n n n      sin(π/n) ln n sin(π/n) 40. 1: lim ln n = lim [sin(π/n) ln n] = lim = 0, so nsin(π/n) → 1 n→∞ n→∞ n→∞ 1/n n

39. 1 :

41. 0 : 42. 1: 43.

ln

n2 + n

=

x3 =0 x→∞ ex √ √ √ √ √ ln( n − 1) 1/ n √ take log: ln( n − 1) = → 0, so ( n − 1)1/ n → 1 n 0≤

n2 ln n n3 < n, n e e

lim (sin x)x = 1

x→0

lim

44.

lim (tan x)tan 2x =

x→π/4

1 e

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SECTION 11.6 

45.

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lim

x→0

1 1 − sin x tan x

 =0

46.

47.

lim (sinh x)−x = 1

x→0+

48.

vertical asymptote x = 1 horizontal asymptote y = 1

vertical asymptote y-axis 49.

50.

horizontal asymptote x-axis

horizontal asymptote x-axis

51.

52.

horizontal asymptote x-axis

vertical asymptote y-axis horizontal asymptote x-axis

53.

√    b 2 b x2 − a2 + x b  2 −ab 2 x −a − x= √ x − a2 − x = √ →0 2 2 2 a a a x −a +x x − a2 + x

54. cosh x − sinh x = 55. for instance, 56. for instance, 57.

lim+ −

x→0

1 x 1 (e + e−x ) − (ex − e−x ) = e−x → 0, 2 2

f (x) = x2 + F (x) = x +

as

x→∞

as x → ∞

(x − 1)(x − 2) x3

sin x 1 + x2

2x 2 = lim . L’Hospital’s rule does not apply here since + cos x x→0 − sin x

lim cos x = 1.

x→0+

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SECTION 11.6 58. (a) Let S be the set of positive integers for which the statement is true. Since lim

x→∞

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

x→∞

(ln x)k+1  (k + 1)(ln x)k = lim =0 x→∞ x x

ln x = 0, x

k ∈ S).

(since

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, (ln x)α (ln x)k < x x

0<

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

 ln a1/x + b1/x = 2

a1/x + b1/x 2 1/x

 .

Now,  ln lim

x→∞

a1/x + b1/x 2 1/x



2 −a1/x ln a − b1/x ln b · 1/x + b1/x  2x2 = lim a x→∞ −1/x2 a1/x ln a + b1/x ln b = x→∞ a1/x + b1/x

= lim

1 2

√ √ Thus, lim ln y = ln ab =⇒ lim y = ab. x→∞

x→∞

  mg 1 − e−(k/m)t  gte−(k/m)t 60. (a) lim v(t) = lim = gt = lim k 1 k→0+ k→0+ k→0+ (b)

dv = g =⇒ v(t) = gt + C; dt

v(0) = 0 =⇒ C = 0

61. (a) Ab = 1 − (1 + b)e−b   2 − 2 + 2b + b2 e−b (b) xb = ; 1 − (1 + b)e−b (c) lim Ab = 1; b→∞

lim xb = 2;

b→∞

yb = lim y b =

b→∞

 1 1 − 1 + 2b + 2b2 e−2b 4 4     (b) Vy = 2π 2 − 2 + 2b + b2 e−b

62. (a) Vx = π

(c) lim Vx = b→∞

π , 4

lim Vy = 4π

b→∞



1 4

−

1 8

 1 + 2b + 2b2 e−2b 2 [1 − (1 + b)e−b ] 1 4



and v(t) = gt.

√ ln ab = ln ab

615 1 ∈ S.

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SECTION 11.6

63. (a)

y

lim

x→0+

3



1 + x2

1/x

=1

g 2 f 1 1

(b)

lim

x→0+



1 + x2

x

2

1/x

=1

since

lim ln



x→0+

 2 1/x

  ln 1 + x2  2x = lim =0 = lim + + x x→0 x→0 (1 + x2 )



1+x

lim f (x) ∼ = 1.5

y

64. (a)

x→∞

1.5

5

10

(b) lim f (x) = lim x→∞

15



x→∞

= lim √ x→∞

65. (a)

20

x

 x2 + 3x + 1 − x

x2

3x + 1 3x + 1 3 = = lim 2 x→∞ 2 + 3x + 1 + x x 1 + (3/x) + 1/x + x lim g(x) ∼ = − 1.7.

y 10

50

x→∞

x

g

-2

(b) lim g(x) = lim x→∞

 3

x→∞

= lim  √ 3 x→∞ =−

 x3 − 5x2 + 2x + 1 − x

−5x2 + 2x + 1 √ 2 x3 − 5x2 + 2x + 1 + x 3 x3 − 5x2 + 2x + 1 + x2

5∼ = −1.667 3

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

+

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

x[P (x)](n−2)/n

b1 xn−1 + b2 xn−2 + · · · + bn b1 as x → ∞ → n [P (x)](n−1)/n + x[P (x)](n−2)/n + · · · + xn−2 [P (x)]1/n + xn−1

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SECTION 11.7 SECTION 11.7

b   ∞  b 1 1 dx dx 1. 1 : =1 = lim = lim − = lim 1 − b→∞ 1 x2 b→∞ x2 x 1 b→∞ b 1 2.

3.

4.



π : 2

∞

dx π = lim tan−1 b = b→∞ 1 + x2 2

∞

dx = lim b→∞ 4 + x2

0



π : 4

0



1 : p

0



0



8

0

dx √ = lim x a→0+



1



1

√

0



1

√

10. 2: 0



2

√

11. 2: 0

 a



b = lim

b→∞

0

a→0





a

1



x dx = lim− b→2 4 − x2

b

√

0

√

a



a→0

  dx 1 = lim =∞ −1 + x2 a a→0+

1

dx = lim− b→1 1 − x2

a

dx π = lim− sin−1 b = b→1 2 1 − x2

√  √ 1 dx = lim −2 1 − x a = lim (0 + 2 1 − a) = 2 a→0+ 1 − x a→0+

b

   −1/2 1/2 b x 4 − x2 dx = lim− − 4 − x2 b→2

0



= lim− 2 − b→2

12.

π : 2



a

0

√

dx = lim− b→a a2 − x2 

∞

13. diverges: e

 14. diverges: e

∞

 1  pb e −1 =∞ p

 8   x−2/3 dx = lim+ 3x1/3 = lim+ 6 − 3a1/3 = 6

8

a

dx = lim 1 − x a→0+

0

1 px e dx = lim b→∞ p

√ dx √ = lim (2 − 2 a) = 2 x a→0+

1

dx = lim x2 a→0+

0

π : 2

e

b→∞

8. diverges:

9.

px

e dx = lim

dx = lim+ a→0 x2/3

7. 6:

b

px

0

6. 2 :

0



∞

5. diverges: 1

  b 1 b dx 1 π −1 x −1 = lim = lim tan tan = b→∞ 2 4 + x2 2 0 b→∞ 2 2 4

b

 −pb  e 1 1 e−px dx = lim − + = b→0 p p p

∞







b

0

ln x dx = lim b→∞ x

dx = lim x ln x b→∞

√

 e

4 − b2



0

=2

   dx b π = lim− sin−1 − sin−1 (0) = b→a a 2 a2 − x2

 e b



b

b

 1 1 ln x 1 2 2 = lim dx = lim (ln x) (ln b) − =∞ b→∞ 2 b→∞ 2 x 2 e

dx = lim [ln(ln x)]be = ∞ x ln x b→∞

617

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December 2, 2006

SECTION 11.7

15. −

1 : 4





1

x ln x dx = lim

x ln x dx = lim

a→0+

0

1

a

∧

a→0+

1 2 1 x ln x − x2 2 4

1 a

(by parts) 

1 1 2 1 2 1 = lim a − a ln a − =− 2 4 4 a→0+ 4



t→0+ ∞

t→0

dx = lim b→∞ x(ln x)2

16. 1: e

 17. π:

18.

∞

−∞

b

e

b  1 1 dx − − = lim = lim + 1 =1 b→∞ x(ln x)2 ln x e b→∞ ln b  b dx dx + lim 2 b→∞ 0 1 + x2 a 1+x  π π b  −1 0  + =π = lim tan x a + lim tan−1 x 0 = − − a→−∞ b→∞ 2 2



19. diverges:

∞



and,

1

lim

c→0+

c



3

20. 2: 1/3



21. ln 2:

dx = lim x2 − 1 b→∞

0

−∞

√ 3





b

−1

dx = lim 3x − 1 a→ 13 +



3

a

3 3(3x − 1)2/3 dx = lim + 2·3 (3x − 1)1/3 a→ 13 a

2/3  8 (3a − 1)2/3 = lim − =2 + 2 2 a→ 13

 1 1 − dx x x+1 1



 b    x b 1 = lim ln = lim ln − ln = ln 2 b→∞ x + 1 1 b→∞ b+1 2

∞

dx = lim x(x + 1) b→∞

1



0

 b  1  d dx dx dx dx + lim− + lim+ + lim ; 2 2 2 2 d→∞ b→0 x x x c→0 a −1 c 1 x

1  1 1 dx = lim+ − = lim −1 =∞ x2 x c c→0+ c c→0

dx = lim a→−∞ x2

−∞



 b

dx 1  x − 1  = lim ln x + 1 2 b→∞ 2 2 x −1 2

   1 b−1 1 1 ln + ln 3 = ln 3 = lim b→∞ 2 b+1 2 2

∞

2



 x

xe dx = lim

a→−∞

 23. 4:



dx = lim a→−∞ 1 + x2

ln 3 : 2

22. −1 :

ln t  1/t 1 = lim = − lim+ t2 = 0. 1/t2 t→0+ −2/t3 2 t→0

lim t2 ln t = lim +

Note:

3

5

√

a

b

xex dx = lim [xex − ex ]0a = lim [−1 − aea + ea ] = −1 a→−∞



x x2

0



−9

dx = lim− a→3

= lim− a→3

5

a→−∞

 −1/2 x x2 − 9 dx

a



 1/2 5 1/2   = lim− 4 − a2 − 9 x2 − 9 =4 a

a→3

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SECTION 11.7 

24.

4

1

 25.

3

−3

619

 b  4

 4 1  x − 2  1  x − 2  dx dx + lim + lim = lim ln ln x + 2 x + 2 2 b→2− 4 a→2+ a x2 − 4 a→2+ 4 1 x −4 1 a            1 b − 2 1 1 1 2 1 a − 2 = lim− ln − ln + lim+ ln − ln = ∞ diverges b→2 4 b + 2 4 3 4 6 4 a + 2 a→2

dx = lim 2 x − 4 b→2−



b



dx x(x + 1)

3

dx diverges: x(x + 1) 0   3  3 1 dx 1 = lim − dx = lim ln |x| − ln |x + 1| a x(x + 1) a→0+ a x x + 1 a→0+

diverges since 

3

0

= lim [ ln 3 − ln 4 − ln a + ln (a + 1)] = ∞. a→0+

26.

 27.



1 : 4 1

−3

∞

1

x dx = lim b→∞ (1 + x2 )2

 1



dx x2 − 4

diverges since 

1

−2

b  −1 −1 x 1 1 dx = lim = lim + = b→∞ 2(1 + x2 ) 1 b→∞ 2(1 + b2 ) (1 + x2 )2 4 4

b

1

−2

dx x2 − 4

dx = lim x2 − 4 a→−2+

diverges: 

= lim

a→−2+

= lim + a→−2

28.

π : 2



∞

0



1 dx = lim x b→∞ e + e−x

0

−∞





29. diverges:  30. Since 1

diverges,

31.

1 : 2

b

cosh x dx = lim

b→∞

0 2

a

1 1 (ln |x − 2| − ln |x + 2|) 4 a

1 [− ln 3 − ln |a − 2| + ln |a + 2| ] = −∞. 4



 b 1 π π −1 x tan dx = lim e = − x −x b→∞ 0 e +e 2 4

b

0

1 dx = lim b→−∞ ex + e−x

∞

 1 1 1 − dx 4 x−2 x+2

1

 b

0

 0 1 π −1 x tan dx = lim e = b→−∞ b ex + e−x 4

cosh x dx = lim [sinh x]b0 = ∞ b→∞

0

  

    b  x − 3 b  dx dx  = lim ln  b − 3  − ln 2  = lim = lim ln x − 2 b − 2 b→2− x2 − 5x + 6 b→2− 1 (x − 2)(x − 3) b→2− 1  4 dx so does 2 1 x − 5x + 6



∞

−x

e 0

 sin x dx = lim

b→∞

0

b

e−x sin x dx = lim − b→∞

b 1  −x e cos x + e−x sin x 0 2

∧ (by parts)  1 1 = lim 1 − e−b cos b − e−b sin b = b→∞ 2 2

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SECTION 11.7 

∞

32. diverges:

cos2 x dx = lim

b→∞

0



1

2e − 2 :

33.

0



π/2

34. 2 : 0

∞

35. (a) converges: 0



∞

(c) converges: 0



2

(c) converges: 0



1

0

π/2

a

1

a

 = lim

b→∞

0

b sin 2b + 2 4



√

 √ π/2 cos x √ dx = lim 2 sin x =2 a a→0+ sin x 

1 x dx = (16 + x2 )2 32

(b) converges:

x π dx = 4 16 + x 16

(d) diverges

 1 sin−1 x dx = x sin−1 x 0 −



1

=∞

 √ 1 e x √ dx = lim 2 e x = 2(e − 1) a x a→0+

√

0

∞

0



√ 3

0







b

√ x3 243 3 4 dx = 55 2−x √ x 8 2 √ dx = 3 2−x

2

36. (a) converges:

37.

√

e x √ dx = lim x a→0+

cos x √ dx = lim a→0+ sin x 

x sin 2x + 2 4

2

(b) converges:

π x2 dx = (16 + x2 )2 16

√

√ 1 dx = 2 2 2−x

√

1 dx = π 2x − x2

0

 (d) converges: 0

x π dx = − lim− 2 2 a→1 1−x

 0

a

√

2

x dx 1 − x2

(by parts)  a  2  √ 1−a2 x 1 1−a 1 √ √ du = − u 1 Now, dx = − = 1 − 1 − a2 2 1 u 1 − x2 0 

u = 1 − x2 1

Thus,

sin−1 x dx =

0



∞

38. (a)

xr e−x dx diverges if r ≤ −1:

0



∞

  π π − lim− 1 − 1 − a2 = − 1. 2 a→1 2

xr e−x dx =



0

1

xr e−x dx +



0

∞



xr e−x dx

1

and

1

xr e−x dx diverges.

0

For any r > −1, we can find k such that xr < ex/2 for x ≥ k (ex/2 grows faster than any power  ∞  k  ∞  ∞ r −x r −x −x/2 of x). Then x e dx < x e dx + e dx, which converges. Thus xr e−x dx 0

0

k

0

converges for all r > −1.  ∞  b   (b) For n = 1 : xe−x dx = lim −xe−x − e−x 0 = lim −be−b − e−b + 1 = 1 b→∞

0



Assume true for n.

∞

n+1 −x

x 0

= lim (−b b→∞

n+1 −b

b→∞

e



e ) + (n + 1) 0

∞

 dx = lim

b→∞

 n+1 −x b −x e 0 + (n + 1)

xn e−x dx = 0 + (n + 1)n! = (n + 1)!



b

n −x

x e 0

 dx

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SECTION 11.7  39.

∞

√

0

 ∞ 1 1 √ dx + dx x (1 + x) x (1 + x) 0 1  1  b 1 1 √ √ = lim dx + lim dx b→∞ 1 x (1 + x) x (1 + x) a→0+ a

1 dx = x (1 + x)





621

1

√

 √ 1 2 √ Now, dx = du = 2 arctan u + C = 2 arctan x + C. 1 + u2 x (1 + x) √ u= x  1   √  √ 1 1 π √ Therefore, lim dx = lim+ 2 arctan x a = lim+ 2 π/4 − arctan a ] = + 2 x (1 + x) a→0 a→0 a→0 a  b   √  √ b 1 π √ and lim dx = lim 2 arctan x 1 = lim 2 arctan b − π/4 = . b→∞ 1 b→∞ b→∞ 2 x (1 + x)  ∞ 1 √ Thus, dx = π. x (1 + x) 0 

∞

40. 1

 b 1 1 √ √ dx + lim dx 2 2 b→∞ a x x −1 2 x x −1 2 ∞   = lim sec−1 x a + lim sec−1 x 2

1 √ dx = lim a→1+ x x2 − 1



2

a→1+

b→∞

= lim (sec−1 2 − sec−1 a) + lim (sec−1 b − sec−1 2) a→1+

b→∞

π π = (sec−1 2 − 0) + ( − sec−1 2) = 2 2

√  ∞  ∞  ∞   1 1 x4 + 1 1 2π 1 + 4 dx = 2π dx = ∞ by comparison with 41. surface area S = dx 3 x x x x 1 1 1 

π/2

42. A =

 (sec x − tan x) dx = lim − b→π/2

0

0

b

 b (sec x − tan x) dx = lim − ln(sec x − tan x) − ln sec x

0

b→π/2

 b = lim − ln(1 + sin x) = ln 2 0

b→π/2



1

1 √ dx = lim x a→0+ 0  √ 1 = lim 2 x a = 2

43. (a)



(b) A =

a

1

1 √ dx x

a→0+

 (c) V =



1

π 0

1 √ x

2

 dx = π 0

1

1 dx = π lim x a→0+

 a

1

1 dx = π lim [ln x]1a x a→0+

diverges

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SECTION 11.7 

44. (a)

∞

1 π dx = lim tan−1 b = b→∞ 1 + x2 2

(b) A = 0

b

1 x π −1 dx = lim x + tan b→∞ 2 (1 + x2 )2 1 + x2 0 0   π b π2 tan−1 b + = lim − 0 = b→∞ 2 1 + b2 4  ∞  b 2πx (d) Vy = dx = lim π ln(1 + x2 ) 0 = ∞ 2 b→∞ 1 + x 0 

∞

(c) Vx =

π·

 45. (a)

∞

(b) A =

e−x dx = 1

0

 (c) Vx =

∞

πe−2x dx = π/2

0

 (d)

∞

Vy =

−x

2πxe

 b→∞

0





∞

(e) A = 

2πe−x



1 + e−2x

0

0

b

2πe−x



0

 b 2πxe−x dx = lim 2π(−x − 1)e−x 0 b→∞

(by parts)

 b+1 = 2π 1 − lim = 2π(1 − 0) = 2π b→∞ eb  b dx = lim 2πe−x 1 + e−2x dx b→∞

 1 + e−2x dx = −2π u = e−x

b

dx = lim

0 e−b



1 + u2 du

1

e−b   = −π u 1 + u2 + ln u + 1 + u2 1  √   √  −b −2b =π 2 + ln 1 + 2 − e 1+e − ln e−b + 1 + e−2b

Taking the limit of this last expression as b → ∞, we have √  √  A=π 2 + ln 1 + 2 . 

∞

xA =1 A 0  ∞ 1 −2x 1 yA 1 yA = dx = , y = e = ; centroid: (1, 14 ) 2 4 A 4 0 1 Yes: 2πxA = 2π = Vy , 2πyA = π = Vx 2

46. xA =

xe−x dx = 1,

x=

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SECTION 11.7



∞

(b) Vy =

−x2

2πxe

2

b

dx = lim

∞



is finite.

b→∞

0

b→∞

0





∞

(c) Vy =

e−x dx

    2 2 b 2 2πxe−x dx = lim π −e−x = lim π 1 − e−b = π

 1 x 1 48. (a) A = − 2 dx = ln 2 x x + 1 2 1  2   ∞  ∞  2  1 x dx (b) Vx = dx < − 2+1 x x x2 1 1 

∞

and 1

 b→∞

0



e−x ≤ e−x

47. (a) The interval [0, 1] causes no problem. For x ≥ 1,

623

2πx 1



x 1 − x x2 + 1



∞

dx = 2π 1

finite

1 1 dx = π 2 x2 + 1 2 

49. (a)

1

(b) A = lim

x

a→0+

dx = lim

a→0+

a

 (c) Vx = lim + a→0

1

1

4 3

= a

 1 πx−1/2 dx = lim+ 2πx1/2 = 2π a

1

(d) Vy = lim

3/4

2πx

a→0+

4 3/4 x 3

a→0

a





−1/4

dx = lim

a→0+

a



∞

8π 7/4 x 7

1 = a

8 π 7

x

g(x) dx = L. Since f (x) ≥ 0 for x ∈ [a, ∞), f (t) dt is increasing. a a  x Therefore it is sufficient to show that f (t) dt is bounded above. For any number

50. (i) Suppose that

M ≥ a, we have

 Therefore,  (ii) If

a





M

M

f (x) dx ≤

a

 g(x) dx ≤

a



x

f (t) dt is bounded and a



∞

f (x) dx diverges,

then

g(x) dx = L a

∞

f (x) dx converges. a

∞

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

0

0



∞

dx x3/2

51. converges by comparison with 1

 52. converges by comparison with



∞

Converges by comparison with π

 56. Diverges by comparison with e

∞

dx x2

∞

e−x dx on [2, ∞).

2

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

54.

∞

1 and x

 1

∞

dx diverges x 

55.

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

converges by comparison with 1

∞

dx x3/2

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SECTION 11.7 

   b 2x 2 ln 1 + x dx = lim =∞ b→∞ 0 1 + x2 b→∞ 0  ∞ 2x Thus, the improper integral dx diverges. 1 + x2 0  b    b (b) 2x 2 lim ln 1 + x dx = lim b→∞ −b 1 + x2 b→∞ −b      = lim ln 1 + b2 − ln 1 + (−b)2 = lim (0) = 0 b

57. (a) lim

b→∞

 58.



∞

sin x dx = lim

b→∞

0

b→∞

b→∞

0

 lim



b

sin x dx = lim



b

−b

sin x dx = lim



L= ∧ (10.7.3)

θ1

b − cos x = 1 − lim cos b; the limit does not exist b→∞

0

b − cos x = lim [− cos b + cos(−b)] = lim [ 0 ] = 0 −b

b→∞

r (θ) = acecθ

59. r(θ) = aecθ ,

b→∞



−∞

b→∞

b→∞

a2 e2cθ + a2 c2 e2cθ dθ

   = a 1 + c2 lim b→−∞

  = a 1 + c2



θ1

b

lim

b→−∞

ecθ c

 ecθ dθ θ1  b

 √     √ 2  cθ  a 1 + c2 1 + c a = lim e 1 − ecb = ecθ1 b→−∞ c c 60. For all real t, − 

∞

61. F (s) =

−sx

e

b→∞

1 ; s

∞

−sx

xe 0

= 

∞

−sx

e 0

x

−∞

e−t

2

 /2

dt

converges by comparison with

b 1 1 −sx dx = lim − e = b→∞ s s 0

x

−∞

et+1 dt.

provided s > 0.

dom(F ) = (0, ∞).

F (s) =

63. F (s) =

b

· 1 dx = lim

 62.

Therefore 

0

Thus, F (s) =



t2 < t + 1. 2

1 s2

e−sx −xe−sx dx = lim − 2 b→∞ s s

if s > 0,

e−sx cos 2x dx = lim

0

Using integration by parts

b→∞



diverges for s ≤ 0, 

b

 = lim

0

b→∞

1 e−sb −be−sb − 2 + 2 s s s



so dom(F ) = (0, ∞).

e−sx cos 2x dx

0 −sx

e

b



4 s −sx 1 −sx cos 2x dx = 2 sin 2x − e cos 2x + C. e s +4 2 4

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SECTION 11.7

625

Therefore,

b 4 1 −sx s −sx F (s) = lim 2 e sin 2x − e cos 2x b→∞ s + 4 2 4 0

 1 −sb 4 s −sb s 4 s s = 2 lim e sin 2b − e cos 2b + = 2 · = 2 s + 4 b→∞ 2 4 4 s +4 4 s +4 Thus,

F (s) =

s ; s2 + 4

dom(F ) = (0, ∞). 

∞

F (s) =

64.



eax e−sx dx =

0

∞

e(a−s)x dx = lim

b→∞

0

1 s−a

=

diverges if s ≤ a;

if s > a,

65. The function f is nonnegative on (−∞, ∞) and  ∞  0  ∞ f (x) dx = 0 dx + −∞



provided s > 0.

−∞

0

e(a−s)b 1 − a−s a−s



so dom F = (a, ∞)

6x dx = (1 + 3x2 )2



∞

0

6x dx (1 + 3x2 )2

6x 1 2 dx = − 1 + 3x2 + C. 2 (1 + 3x )

Now, Therefore,



f (x) dx = lim −

∞

1 1 + 3x2

b→∞

−∞

b

 = lim

b→∞

0

1 1− 1 + 3b2

 = 1.

66. f is nonnegative, and  ∞  ∞   b  f (x) dx = ke−kx dx = lim −e−kx 0 = lim −e−kb + 1 = 1 −∞

b→∞

0

b→∞

So, f is a probability density function.  67.

μ=



∞

−∞

xf (x) dx =



0

−∞

∞

kxe−kx dx = lim

0 dx + 

Using integration by parts,



b→∞

0

b

kxe−kx dx

0

kxe−kx dx = −xe−kx −

1 −kx e + C. k

Therefore,

b

  ∞ 1 1 1 1 μ= = xf (x) dx = lim −xe−kx − e−kx = lim −be−kb − e−kb + b→∞ b→∞ k k k k −∞ 0  68. σ =

∞

−∞

 =

∞

 (x − μ)2 f (x) dx = kx2 e−kx dx − 2

1 k

x−

∞

1 k

ke−kx dx 

∞

e−kx dx =

0

1 k2 

t

F (t) =

f (x) dx 1

is increasing, and that

2

xe−kx dx +

0

 69. Observe that



0



0

∞

(∗) an ≤

is continuous and increasing, that

t 1

f (x) dx ≤ an+1

an =

for t ∈ [ n, n + 1 ].

n

f (x) dx 1

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REVIEW EXERCISES  If

∞

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

f (x) dx 1

and therefore convergent. If {an } converges, then {an } is bounded and, by (∗), F is bounded. Being  ∞ increasing, F is also convergent; i.e., f (x) dx converges. 1

REVIEW EXERCISES 1. |x − 2| ≤ 3 =⇒ −1 ≤ x ≤ 5 : 2. x2 > 3 =⇒ x >

√

lub = 5, glb = −1.

√ 3 or x < − 3;

no lub, no glb.

3. x2 − x − 2 ≤ 0 =⇒ (x − 2)(x + 1) ≤ 0 =⇒ −1 ≤ x ≤ 2 : 4. cos x ≤ 1 for all x;

lub = 2, glb = −1.

no lub, no glb.

5. Since e−x ≤ 1 for all x, 2

e−x ≤ 2 for all x; 2

6. ln x < e =⇒ 0 < x < ee :

no lub, no glb.

lub = ee , glb = 0.

7. increasing; bounded below by

1 2

and above by 23 .

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

n2 − 1 1 = n − → ∞ as n → ∞. n n

9. bounded below by 0 and above by 32 ; not monotonic 10. increasing; bounded below by 11.

12.

4 5

and above by 1.

! # " 2n , . . . ; the sequence is not monotonic. = 2, 1, 89 , 1, 32 25 2 n However, it is increasing from a3 on. The sequence is bounded below by 89 ; it is not bounded above. sin (nπ/2) n2

!

" = 1, 0, − 19 , 0,

#

1 25 , · · ·

;

bounded below by − 19 and above by 1; not monotonic

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

14. converges to 1:

3 1+ + n2 + 3n + 2 n = 7 n2 + 7n + 12 1+ + n

15. converges to 1:

1 lim ln n→∞ n



n n+1

2 n2 → 1. 12 n2 

 = 0 =⇒ lim

n→∞

n 1+n

1/n =1

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REVIEW EXERCISES 16. converges to 0:

5 4 1 + 2+ 3 4n2 + 5n + 1 n n n = → 0. 1 n3 + 1 1+ 3 n

17. converges to 0:

cos nπ sin nπ → cos 0 sin 0 = 0 as n → ∞

627

18. diverges: (2 + n1 )n > 2n and 2n diverges. 0 = [ln 1]n ≤ [ln(1 +

19. converges to 0:

3 ln 2n − ln(n3 + 1) = ln

20. converges to ln 8:

3 2

21. converges to

1 n ) ] ≤ [ln 2]n ; n

√

:

[ln 2]n → 0 as n → ∞.

8n3 → ln 8. n3 + 1

3 − n12 3n2 − 1 = 4n4 + 2n2 + 3 4 + n22 +

3 n4

→

3 as n → ∞. 2

1

(n2 + 4) 3 (1/n + 4/n3 )1/3 = → 0 as n → ∞. 2n + 1 2 + 1/n

22. converges to 0:

π π cos → 0 cos 0 = 0 as n → ∞. n n

23. converges to 0: 24. converges 0:

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

n+1

25. converges to 0: n

 26. diverges: 1

n

n+1  1 e−x dx = − e−x = e−n (1 − ) → 0 as n → ∞ n e

 √ n √ √ 1 √ dx = 2 x = 2 n − 2 and 2 n − 2 diverges. 1 x

27. Given  > 0. Since an → L, there exists a positive integer K such that if n ≥ K, then |an − L| < . Now, if n ≥ K − 1, then n + 1 ≥ K and |an+1 − L| < . Therefore, an+1 → L. 28. Let  > 0. Since an → L, there is positive integer K such that if n ≥ K, |an − L| <

 . 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|  < , i = 1, 2, · · · , K. n 2K Let M = max {K, N }. Then, if n ≥ M ,   K n   a1 + · · · + an   |ai − L| |aj − L|    ≤ − L + < K( ) + n( ) = .   n n n 2K 2n i=1 j=K+1

Therefore mn → L.

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REVIEW EXERCISES

29. As an example, let a =

π 3.

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 (cos x) 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. ln x 5+2 5x + 2 ln x x = 5; 31. lim = lim ln x x→∞ x + 3 ln x x→∞ 1+3 x 32.



 ln x =0 lim x→∞ x

1 ex − 1  ex = = lim 2 x→0 tan 2x x→0 2 sec 2x 2 lim

− sin x ln(cos x)  −1 sin x 1 33. lim = lim cos x = lim · =− 2 x→0 x→0 x→0 2 cos x x 2x x 2 34. Set y = x1/(x−1) . Then ln y =

ln x , x−1

and

lim

x→1

ln x  1 = lim = 1 x − 1 x→1 x

Therefore, lim x1/(x−1) = e. x→1

35.

36.

lim (1 +

x→∞

2 4 2x 4 ) = lim (1 + )x = e8 x→∞ x x

e2x − e−2x  2e2x + 2e−2x = lim =4 x→0 x→0 sin x cos x lim

1 2 ln x  x = lim −x = 0 37. lim x2 ln x = lim = lim x→0+ x→0+ 1 x→0+ −2 x→0+ 2 x2 x3 38.

39.

10x  10x ln 10  10x (ln 10)2  10x (ln 10)10 = lim = lim = · · · = lim →∞ 10 9 8 x→∞ x x→∞ x→∞ (10)(9)x x→∞ 10x 10! lim

ex + e−x − x2 − 2  ex − e−x − 2x  ex − e−x − 2  ex + e−x − 2 = lim = lim = lim x→0 x→0 2 sin x cos x − 2x x→0 sin 2x − 2x x→0 2 cos 2x − 2 sin2 x − x2 lim



ex − e−x  ex + e−x 1 = lim =− x→0 −4 sin x2x x→0 −8 cos 2x 4

= lim

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REVIEW EXERCISES ln x  1/x 1 = lim =− sin πx x→1 π cos πx π  x 2  x 2 et dt ex x2 1 2 2  41. lim xe−x = et dt = lim 0 x2 = lim 2 2 = lim 2 2 x x x→∞ x→∞ x→∞ 2x e x→∞ 2x − 1 2 e −e 0 2 x x

40.

lim csc(πx) ln x = lim

x→1

x→1



e−1/x 42. Consider ln = xn 2

 1 − 2 − n ln |x| . x   1 1 + nx2 ln |x| lim − 2 − n ln |x| = lim − = −∞ x→0 x→0 x x2

e−1/x → 0. x→0 xn 2

Therefore 

∞

43.

√

e− √

1

 √

44.



1

45. 0

lim



x

x

b

dx = lim

b→∞

√

e− √

1



x

x

dx = lim

b→∞

√

− 2e−

x

b 1

√

= lim (−2e−

b

b→∞

+ 2e−1 ) = 2e−1

x dx = − 1 − x2 + C 1 − x2  1  b  b x x √ √ dx = lim− dx = lim− − 1 − x2 = lim− (− 1 − b2 + 1) = 1 b→1 b→1 b→1 0 1 − x2 1 − x2 0 0   a 1 1 1 1 − dx (partial fractions) dx = − lim− 2 2 a→1 0 x−1 x+1 0 1−x 

a  1 1 a+1 = lim− ln(x + 1) − ln(1 − x) = lim− ln = ∞; 0 2 a→1 2 a→1 1−a

1 dx = lim− a→1 1 − x2



a

the integral diverges. 



π/2

46.

sec x dx = lim − c→π/2

0

c

0

 c sec x dx = lim − ln(sec x + tan x) = lim − ln(sec c + tan c) = ∞; c→π/2

0

the integral diverges. 

∞

47. 1

sin(π/x) dx = lim b→∞ x2

 1

b

1 b sin(π/x) 2 dx = lim = cos π/x b→∞ π 1 x2 π

 c  9 1 1 1 48. dx = lim− dx + lim+ dx 2/3 2/3 2/3 c→1 c→1 0 (x − 1) 0 (x − 1) c (x − 1)  c  c 1 1/3 lim− dx = lim 3 (x − 1) =3 2/3 c→1 c→1− 0 0 (x − 1)  9 9  1 1/3 lim dx = lim 3 (x − 1) =6 c c→1+ c (x − 1)2/3 c→1+  9 1 = 3 + 6 = 9. (x − 1)2/3 0 

9

c→π/2

629

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December 2, 2006

REVIEW EXERCISES 

49. 

∞

0

1 ex + e−x



ex dx = arctan ex + C +1  c c 1 π  dx = lim dx = lim arctan ex  = x −x c→∞ 0 e + e c→∞ 0 2

1 dx = ex + e−x

e2x

50. Set u = ln x, du =  2

∞

1 dx = x(ln x)k



1 dx; u(2) = ln 2. Then x ∞

ln 2

1 du uk

The integral converges if k > 1 and diverges otherwise. For k > 1,



∞

2



51.

1 dx = lim c→∞ x(ln x)k 

a

a

ln(1/x) dx = 0

0



c

ln 2

c 1 1 1  du = lim = u1−k  k c→∞ 1 − k ln 2 u (k − 1) (ln 2)k−1 

− ln x dx = lim+ c→0

c

a

 a − ln x dx = lim+ − x ln x + x c→0

c

= lim+ [−a ln a + a + c ln c − c] = a ln(1/a) + a c→0

52. y = (a2/3 − x2/3 )3/2 ; y  = −x−1/3 (a2/3 − x2/3 )1/2  a  a 3a L= 1 + (y  )2 dx = a1/3 x−1/3 dx = 2 0 0 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  = (lub(S) + lub(T )) − M . There exist s ∈ S and t ∈ T such that lub(S) − s < /2,

and

lub(T ) − t < /2

Now, lub(S) + lub(T ) − (s + t) = lub(S) − s + lub(T ) − t <  = 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).

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P1: PBU/OVY JWDD027-11

P2: PBU/OVY

QC: PBU/OVY

T1: PBU

JWDD027-Salas-v1

December 2, 2006

REVIEW EXERCISES  55. (a) If

∞

−∞

f (x)dx = L, then 



c

lim

c→∞

both exist and

f (x) dx, 

c→∞

and

0

b→−∞

Let c = −b, then

 lim

c→∞

 (b) Set f (x) = x. then lim

c→∞

x dx = 0, but

−c

Now assume that lim

c→∞



lim

f (x) dx = L b

∞

−∞

c→∞

x dx diverges.

c

−c

f (x) dx = L

c

−c

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

x

f (t) dt ≤ L on

[0, ∞).

0 x

Therefore 



0

f (x) dx = L

56. (a) Assume that f is nonnegative on (−∞, ∞).  ∞  By Exercise 55, f (x) dx = L =⇒ lim −∞

f (x) dx b

c



c

−c

b→−∞

f (x) dx + lim 0

0

lim



c

lim

c→∞

631

f (t) dt is a bounded and nondecreasing function, which implies that 0



c

f (x) dx exists. Similarly, lim 0



Therefore,

c→∞

0

−c

f (x) dx exists.

∞

−∞

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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