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= .
3


Chapter 5
limx’0 (cos x ’ 1) = 0 and limx’0 x 2 ’ x 3 = 0 so l™Hôpital™s
1. (a)
Rule applies. Thus

cos x ’ 1 ’ sin x
= lim
lim .
x’0 x 2 ’ x 3 x’0 2x ’ 3x 2
294 Solutions to Exercises

Now l™Hôpital™s Rule applies again to yield
’ cos x 1
= lim =’ .
x’0 2 ’ 6x 2
limx’0 e2x ’ 1 ’ 2x = 0 and limx’0 x 2 + x 4 = 0 so l™Hôpital™s
(b)
Rule applies. Thus
e2x ’ 1 ’ 2x 2e2x ’ 2
= lim
lim .
x2 + x4 x’0 2x + 4x 3
x’0
l™Hôpital™s Rule applies again to yield
4e2x
= lim = 2.
x’0 2 + 12x 2
limx’0 cos x = 0, so l™Hôpital™s Rule does not apply. In fact the
(c)
limit does not exist.
limx’1 [ln x]2 = 0 and limx’1 (x ’ 1) = 0 so l™Hôpital™s Rule
(d)
applies. Thus
[ln x]2 [2 ln x]/x
= lim = 0.
lim
x’1 (x ’ 1) 1
x’1

limx’2 (x ’ 2)3 = 0 and limx’2 sin(x ’ 2) ’ (x ’ 2) = 0 so
(e)
l™Hôpital™s Rule applies. Thus
(x ’ 2)3 3(x ’ 2)2
= lim
lim .
x’2 sin(x ’ 2) ’ (x ’ 2) x’2 cos(x ’ 2) ’ 1

Now l™Hôpital™s Rule applies again to yield
6(x ’ 2)
= lim .
x’2 ’ sin(x ’ 2)

We apply l™Hôpital™s Rule one last time to obtain
6
= lim = ’6.
x’2 ’ cos(x ’ 2)

limx’1 (ex ’ 1) = 0 and limx’1 (x ’ 1) = 0 so l™Hôpital™s Rule
(f)
applies. Thus
ex ’ 1 ex
= lim = e.
lim
x’1 x ’ 1 x’1 1

limx’+∞ x 3 = limx’+∞ (ex ’ x 2 ) = +∞ so l™Hôpital™s Rule
2. (a)
applies. Thus
x3 3x 2
= lim x
lim .
x’+∞ e ’ 2x
x’+∞ ex ’ x 2
295
Chapter 5

l™Hôpital™s Rule applies again to yield
6x
= lim .
x’+∞ ex ’ 2

l™Hôpital™s Rule applies one more time to ¬nally yield
6
= 0.
lim
x’+∞ ex

limx’+∞ ln x = limx’+∞ x = +∞ so l™Hôpital™s Rule applies.
(b)
Thus
ln x 1/x
= lim = 0.
lim
x’+∞ 1
x’+∞ x

limx’+∞ e’x = limx’+∞ ln[x/(x + 1)] = 0 so l™Hôpital™s Rule
(c)
applies. Thus
e’x ’e’x
= lim
lim .
x’+∞ ln[x/(x + 1)] x’+∞ 1/x ’ 1/[x + 1]

It is convenient to rewrite this expression as
x2 + x
lim .
x’+∞ ’ex

Now l™Hôpital™s Rule applies once more to yield
2x + 1
lim .
x’+∞ ’ex

We apply l™Hôpital™s Rule a last time to obtain
2
= lim = 0.
x’+∞ ’ex

(d) Since limx’+∞ sin x does not exist, l™Hôpital™s Rule does not apply.
In fact the requested limit does not exist.
(e) It is convenient to rewrite this limit as
x
lim ’x .
x’’∞ e

Since limx’’∞ x = limx’’∞ e’x = ±∞, l™Hôpital™s Rule
applies. Thus
1
x
= lim = 0.
lim
x’’∞ e’x x’’∞ ’e’x
296 Solutions to Exercises

Since limx’’∞ ln |x| = limx’’∞ e’x = +∞, l™Hôpital™s Rule
(f)
applies. Thus
ln |x| 1/x
= lim = 0.
lim
x’’∞ e’x x’’∞ ’e’x

x3
We write the limit as limx’+∞ x . Since limx’+∞ x 3 = limx’+∞
3. (a)
e
ex = +∞, l™Hôpital™s Rule applies. Thus
x3 3x 2
3 ’x
= lim x = lim
lim x e .
x’+∞ ex
x’+∞ e
x’+∞

We apply l™Hôpital™s Rule again to obtain
6x
= lim .
x’+∞ ex

Applying l™Hôpital™s Rule one last time yields
6
= lim = 0.
x’+∞ ex

sin(1/x)
(b) We write the limit as limx’+∞ . Since limx’+∞ sin(1/x)
1/x
= limx’+∞ 1/x = 0, l™Hôpital™s Rule applies. Hence
sin(1/x)
lim x · sin[1/x] = lim
1/x
x’+∞ x’+∞

[cos(1/x)] · [’1/x 2 ]
= lim
’1/x 2
x’+∞
cos(1/x)
= lim = 1.
1
x’+∞

ln[x/(x + 1)]
(c) We rewrite the limit as limx’+∞ . Since limx’+∞
1/(x + 1)
ln[x/(x + 1)] = limx’+∞ 1/(x + 1) = 0, l™Hôpital™s Rule applies.
Thus
ln[x/(x + 1)]
lim ln[x/(x + 1)] · (x + 1) = lim
x’+∞ 1/(x + 1)
x’+∞

[(x + 1)/x] · [1/(x + 1)2 ]
= lim
’1/(x + 1)2
x’+∞
’(x + 1)
= lim .
x
x’+∞
297
Chapter 5

Now l™Hôpital™s Rule applies again and we obtain
’1
= lim
= ’1.
x’+∞ 1

[ln x]
We rewrite the limit as limx’+∞ x . Since limx’+∞ ln x =
(d)
e
limx’+∞ ex = +∞, l™Hôpital™s Rule applies. Thus
ln x 1/x
lim ln x · e’x = lim = lim = 0.
x’+∞ ex x’+∞ ex
x’+∞

x2
. Since limx’’∞ lim x 2 =
(e) We write the limit as limx’’∞
e’2x
limx’’∞ e’2x = 0, l™Hôpital™s Rule applies. Thus
x2
2x
· x = lim = lim
2x 2
lim e .
x’’∞ e’2x x’’∞ ’2e’2x
x’’∞

l™Hôpital™s Rule applies one more time to yield
2
= lim = 0.
4e’2x
x’’∞

e1/x
. Since limx’0 e1/x =
(f) We rewrite the limit as limx’0
[1/x]
limx’0 1/x = +∞, l™Hôpital™s Rule applies. Thus
e1/x · [’1/x 2 ]
e1/x e1/x
= lim = lim = lim = +∞ .
1/x
lim x·e
’1/x 2
x’0 1/x x’0 1
x’0 x’0

4. We do (a), (b), (c), (d).
1
1 1 x 1/4
’3/4 ’3/4
dx = lim dx = lim
(a) x x
’0+ ’0+ 1/4
0

11/4 1/4
= lim ’ = 4.
’0+ 1/4 1/4
3 3’
’4/3
(x ’ 3)’4/3 dx
(x ’ 3) dx = lim
(b)
’0+ 1
1
3’
(x ’ 3)’1/3 ’ ’1/3 ’2’1/3
= lim = lim ’ . But
’1/3 ’1/3 ’1/3
’0+ ’0+
1
the limit does not exist; so the integral does not converge.
’1’
2 1 1
dx = lim
(c) dx
(x + 1)1/3 (x + 1)1/3
’0+
’2 ’2
298 Solutions to Exercises

2 1
+ lim dx
(x + 1)1/3
’0+ ’1+
’1’ 2
(x + 1)2/3 (x + 1)2/3
= lim + lim
’0+ ’0+
2/3 2/3
’2 ’1+

(’ )2/3 (’1)2/3 32/3 ( )2/3
= lim ’ + lim ’
’0+ ’0+
2/3 2/3 2/3 2/3
3
= · 32/3 ’ 1 .
2
’2’
6 x x
dx = lim
(d) dx
(x ’ 1)(x + 2) (x ’ 1)(x + 2)
’0+
’4 ’4
0 1’
x x
+ lim dx + lim dx
(x ’1)(x +2) (x ’1)(x +2)
’0+ ’2+ ’0+ 0




Y
6 x
+ lim dx. Now
FL
(x ’ 1)(x + 2)
’0+ 1+

1/3 2/3
x
= + .
AM

(x ’ 1)(x + 2) x’1 x+2
Therefore
6 x
TE



dx
’4 (x ’1)(x +2)
’2’ 1/3 0
2/3 1/3 2/3
= lim + dx + lim + dx
x ’1 x +2 x ’1 x +2
’0+ ’4 ’0+ ’2+
1’ 6
1/3 2/3 1/3 2/3
+ lim + dx + lim + dx
x ’1 x +2 x ’1 x +2
’0+ 0 ’0+ 1+
’2’
1 2
= lim ln |x ’1|+ ln |x +2|
’0+ 3 3 ’4
0
1 2
+ lim ln |x ’1|+ ln |x +2|
’0+ 3 3 ’2+
1’
1 2
+ lim ln |x ’1|+ ln |x +2|
’0+ 3 3 0
6
1 2
+ lim ln |x ’1|+ ln |x +2| .
’0+ 3 3 1+
299
Chapter 5

Now this equals

1 2 1 2
· ln | ’ 3 ’ | + ln ’ · ln 5 + ln 2 + etc.
lim
’0+ 3 3 3 3

The second limit does not exist, so the original integral does not
converge.
5. We do (a), (b), (c), (d).
N
∞ e’3x
’3x ’3x
dx = dx =
(a) lim e lim
e
’3
N ’+∞ N ’+∞
1 1
e’3N e’3 e’3
= ’ =
lim .
’3 ’3
N ’+∞ 3
∞ N
2 ’x
x 2 e’x dx
dx =
(b) lim
xe
N ’+∞ 2
2

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. 34
( : 41)



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