Page 307 - Calculus Demystified

Solutions to Exercises
294
Now l’Hôpital’s Rule applies again to yield
− cos x 1
= lim =− .
x→0 2 − 6x 2
2
4
(b) lim x→0 e 2x − 1 − 2x = 0 and lim x→0 x + x = 0 so l’Hôpital’s
Rule applies. Thus
e 2x − 1 − 2x 2e 2x − 2
lim = lim .
2
x→0 x + x 4 x→0 2x + 4x 3
l’Hôpital’s Rule applies again to yield
4e 2x
= lim = 2.
x→0 2 + 12x 2
(c) lim x→0 cos x = 0, so l’Hôpital’s Rule does not apply. In fact the
limit does not exist.
2
(d) lim x→1 [ln x] = 0 and lim x→1 (x − 1) = 0 so l’Hôpital’s Rule
applies. Thus
2
[ln x] [2ln x]/x
lim = lim = 0.
x→1 (x − 1) x→1 1
3
(e) lim x→2 (x − 2) = 0 and lim x→2 sin(x − 2) − (x − 2) = 0so
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)
x
(f) lim x→1 (e − 1) = 0 and lim x→1 (x − 1) = 0 so l’Hôpital’s Rule
applies. Thus
x
e − 1 e x
lim = lim = e.
x→1 x − 1 x→1 1
2
3
x
2. (a) lim x→+∞ x = lim x→+∞ (e − x ) =+∞ so l’Hôpital’s Rule
applies. Thus
x 3 3x 2
lim = lim .
x
x
x→+∞ e − x 2 x→+∞ e − 2x