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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
<0+
n k 1 1 0≤ <0+ 2 4
1 10
k
k+1 n> 2
for n > 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 <0 x2 1 3
for 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 · <1 = = an (n + 1)! 5n n+1 Sequence is not nonincreasing: 42. For n ≥ M,
a1 = 5 <
25 2
and thus an+1 < an .
= a2 .
an+1 M n+1 M n! = = · < 1, an (n + 1)! M n n+1
so the sequence decreases for n ≥ M .
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SECTION 11.2 43. boundedness:
0 < (cn + dn )1/n < (2dn )1/n = 21/n d ≤ 2d
n+1 monotonicity : an+1 + d n+1 = ccn + ddn n+1 = c
< (cn + dn )1/n cn + (cn + dn )1/n dn = (cn + dn )1+1/n = (cn + dn )(n+1)/n = an+1 n 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 |an bn | = |an ||bn | ≤ M N 45. a1 = 1,
a2 = 12 ,
a3 = 16 ,
a4 =
46. a1 = 1,
a2 = 8,
a3 = 27,
a4 = 64,
47. a1 = a2 = a3 = a4 = a5 = a6 = 1;
1 24 ,
a5 =
1 120 ,
a5 = 125,
a6 =
1 720 ;
a6 = 216;
an =
1 n!
an = n3
an = 1
a2 = 32 ,
a3 = 74 ,
a4 =
49. a1 = 1,
a2 = 3,
a3 = 5,
a4 = 7,
50. a1 = 1,
a2 = 12 ,
a3 = 13 ,
a4 = 14 ,
a5 = 15 ,
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;
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;
57. First a1 = 21 − 1 = 1.
15 8 ,
a5 =
31 16 ,
a5 = 9,
a6 =
63 32 ;
an = (2n − 1)/2n−1
48. a1 = 1,
an = 2n − 1
a6 = 11; a6 = 16 ;
an = 1/n
an = 2n−2
an = n + 1
(n ≥ 3)
(n ≥ 2)
Next suppose ak = 2k − 1 for some k ≥ 1. Then ak+1 = 2ak + 1 = 2 2k − 1 + 1 = 2k+1 − 1.
593
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SECTION 11.2 Then an+1 = an + 5 = 5n − 2 + 5 = 5(n + 1) − 2.
58. True for n = 1. Assume true for n. 59. First a1 =
1 = 1. 20
Next suppose ak = ak+1 =
k
for some
2k−1
k ≥ 1.
Then
k+1 k+1 k k+1 = . ak = k−1 2k 2k 2 2k
60. True for n = 1. Assume true for n. Then an+1 = an −
1 1 1 n+1−1 1 = − = = n(n + 1) n n(n + 1) n(n + 1) n+1
61. (a) If r = 1 then Sn = n for n = 1, 2, 3, . . . Sn = 1 + r + r2 + · · · + rn−1
(b)
rSn = r + r2 + · · · + rn Sn − rSn = 1 − rn 1 − rn Sn = , 1−r 62.
1 1 1 = − , k(k + 1) k k+1
r = 1.
so
Sn = a1 + a2 + · · · + an−1 + an 1 1 1 1 + + ··· + + 1·2 2·3 (n − 1)n n(n + 1) 1 1 1 1 1 1 1 = 1− + − + ··· + − + − 2 2 3 n−1 n n n+1 1 n = 1− = since all middle terms cancel out. n+1 n+1
=
63. (a) Let Sn denote the distance traveled between the nth and (n + 1)st bounce. Then n−1 3 3 3 3 S2 = 4 (75) + 4 (75) = 150 . S1 = 75 + 75 = 150, ,..., Sn = 150 4 4 (b) An object dropped from rest from a height h feet above the ground will hit the ground in seconds. Therefore it follows that the ball will be in the air
√ (n−1)/2 1 Sn 5 3 3 Tn = 2 4 = 2 2 4 64. P0 = 5,
P12 = 5 · 2,
P24 = 5 · 22 ;
65. {an } is an increasing sequence; 66. {an } is a decreasing sequence;
so
an →
1 2
1 4
√
h
seconds.
Pn = 5 2n/12 as n → ∞.
an → 2 as n → ∞.
67. (a) Let S be the set of positive integers for which an+1 > 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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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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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+
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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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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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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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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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