Page 309 - Calculus Demystified

Solutions to Exercises
296
(f) Since lim x→−∞ ln |x|= lim x→−∞ e −x =+∞, l’Hôpital’s Rule
applies. Thus
ln |x| 1/x
lim = lim = 0.
x→−∞ e −x x→−∞ −e −x
x 3
3
3. (a) Wewritethelimitaslim x→+∞ .Sincelim x→+∞ x = lim x→+∞
e x
x
e =+∞, l’Hôpital’s Rule applies. Thus
x 3 3x 2
3 −x
lim x e = lim = lim .
x→+∞ x→+∞ e x x→+∞ e x
We apply l’Hôpital’s Rule again to obtain
6x
= lim .
x→+∞ e x
Applying l’Hôpital’s Rule one last time yields
6
= lim = 0.
x→+∞ e x

sin(1/x)
(b) We write the limit as lim x→+∞ . Since lim x→+∞ sin(1/x)
1/x
= lim x→+∞ 1/x = 0, l’Hôpital’s Rule applies. Hence
sin(1/x)
lim x · sin[1/x]= lim
x→+∞ x→+∞ 1/x
2
[cos(1/x)]·[−1/x ]
= lim
x→+∞ −1/x 2
cos(1/x)
= lim = 1.
x→+∞ 1
ln[x/(x + 1)]
(c) We rewrite the limit as lim x→+∞ . Since lim x→+∞
1/(x + 1)
ln[x/(x +1)]= lim x→+∞ 1/(x +1) = 0, l’Hôpital’s Rule applies.
Thus
ln[x/(x + 1)]
lim ln[x/(x + 1)]· (x + 1) = lim
x→+∞ x→+∞ 1/(x + 1)
2
[(x + 1)/x]·[1/(x + 1) ]
= lim
x→+∞ −1/(x + 1) 2
−(x + 1)
= lim .
x→+∞ x