Page 140 - Calculus Demystified

CHAPTER 5
and then Indeterminate Forms 127
24
lim .
x→+∞ e x
We conclude that
x 4
lim = 0.
x→+∞ e x
e
x
You Try It: Evaluate the limit lim x→+∞ .
x ln x
x
4
You Try It: Evaluate the limit lim x→−∞ x · e .
EXAMPLE 5.5
Evaluate the limit
sin(2/x)
lim .
x→−∞ sin(5/x)
SOLUTION
We note that both numerator and denominator tend to 0, so the quotient is
indeterminate at −∞ of the form 0/0. We may therefore apply l’Hôpital’s Rule.
Our limit equals
2
(−2/x ) cos(2/x)
lim .
2
x→−∞ (−5/x ) cos(5/x)
This in turn simplifies to
2 cos(2/x) 2
lim = .
x→−∞ 5 cos(5/x) 5
l’Hôpital’s Rule also applies to one-sided limits. Here is an example.

EXAMPLE 5.6
Evaluate the limit
√
sin x
lim √ .
x→0 + x
SOLUTION
Both numerator and denominator tend to zero so the quotient is indeterminate
at 0 of the form 0/0. We may apply l’Hôpital’s Rule; differentiating numerator
and denominator, we find that the limit equals
√
[cos x]· (1/2)x −1/2 √
lim = lim cos x
x→0 + (1/2)x −1/2 x→0 +
= 1.
You Try It: How can we apply l’Hôpital’s Rule to evaluate lim x→0 + x · ln x?