Page 308 - Calculus Demystified

Chapter 5
l’Hôpital’s Rule applies again to yield 295
6x
= lim .
x
x→+∞ e − 2
l’Hôpital’s Rule applies one more time to finally yield
6
lim = 0.
x→+∞ e x

(b) lim x→+∞ ln x = lim x→+∞ x =+∞ so l’Hôpital’s Rule applies.
Thus
ln x 1/x
lim = lim = 0.
x→+∞ x x→+∞ 1
(c) lim x→+∞ e −x = lim x→+∞ ln[x/(x + 1)]= 0 so l’Hôpital’s Rule
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
2
x + x
lim .
x→+∞ −e x
Now l’Hôpital’s Rule applies once more to yield

2x + 1
lim .
x→+∞ −e x
We apply l’Hôpital’s Rule a last time to obtain
2
= lim = 0.
x→+∞ −e x
(d) Sincelim x→+∞ sin x doesnotexist,l’Hôpital’sRuledoesnotapply.
In fact the requested limit does not exist.
(e) It is convenient to rewrite this limit as
x
lim .
x→−∞ e −x

Since lim x→−∞ x = lim x→−∞ e −x =±∞, l’Hôpital’s Rule
applies. Thus

x 1
lim = lim = 0.
x→−∞ e −x x→−∞ −e −x