Page 310 - Calculus Demystified

Chapter 5
Now l’Hôpital’s Rule applies again and we obtain 297
−1
= lim =−1.
x→+∞ 1
[ln x]
(d) We rewrite the limit as lim x→+∞ . Since lim x→+∞ ln x =
e x
x
lim x→+∞ e =+∞, l’Hôpital’s Rule applies. Thus
ln x 1/x
−x
lim ln x · e = lim = lim = 0.
x→+∞ x→+∞ e x x→+∞ e x
x 2
2
(e) We write the limit as lim x→−∞ . Since lim x→−∞ lim x =
e −2x
lim x→−∞ e −2x = 0, l’Hôpital’s Rule applies. Thus
x 2 2x
2
lim e 2x · x = lim = lim .
x→−∞ x→−∞ e −2x x→−∞ −2e −2x
l’Hôpital’s Rule applies one more time to yield
2
= lim = 0.
x→−∞ 4e −2x
e 1/x
(f) We rewrite the limit as lim x→0 . Since lim x→0 e 1/x =
[1/x]
lim x→0 1/x =+∞, l’Hôpital’s Rule applies. Thus
2
e 1/x e 1/x ·[−1/x ] e 1/x
lim x·e 1/x = lim = lim = lim =+∞ .
x→0 x→0 1/x x→0 −1/x 2 x→0 1
4. We do (a), (b), (c), (d).
1 1 x 1/4
1
(a) x −3/4 dx = lim x −3/4 dx = lim
0 (→0 + ( (→0 + 1/4 (
1/4 1/4
1 (
= lim − = 4.
(→0 + 1/4 1/4
3 3−(

(b) (x − 3) −4/3 dx = lim (x − 3) −4/3 dx
1 (→0 + 1
3−(
−1/3 −1/3 −1/3
(x − 3) −( −2
= lim = lim − . But
(→0 + −1/3 1 (→0 + −1/3 −1/3
the limit does not exist; so the integral does not converge.
2 1 −1−( 1

(c) dx = lim dx
−2 (x + 1) 1/3 (→0 + −2 (x + 1) 1/3