Page 138 - Calculus Demystified

CHAPTER 5
SOLUTION Indeterminate Forms 125
As x → 0 both numerator and denominator tend to zero, so the quotient is
indeterminate at 0 of the form 0/0. Thus l’Hôpital’s Rule applies. Our limit
equals
(d/dx)x 3
lim ,
x→0 (d/dx)(x − sin x)
provided that this last limit exists. It equals

3x 2
lim .
x→0 1 − cos x
This is another indeterminate form. So we must again apply l’Hôpital’s Rule.
The result is
6x
lim .
x→0 sin x
This is again indeterminate; another application of l’Hôpital’s Rule gives us
finally
6
lim = 6.
x→0 cos x
We conclude that the original limit equals 6.

You Try It: Apply l’Hôpital’s Rule to the limit lim x→0 x/[1/ ln |x|].

Indeterminate Forms Involving ∞ We handle indeterminate forms involv-
ing infinity as follows: Let f(x) and g(x) be differentiable functions on (a, c) ∪
(c, b). If

lim f(x) and lim g(x)
x→c x→c
both exist and equal +∞ or −∞ (they may have the same sign or different signs)
then

f(x) f (x)
lim = lim ,

x→c g(x) x→c g (x)
provided this last limit exists either as a finite or infinite limit.
Let us look at some examples.
EXAMPLE 5.3
Evaluate the limit
3
lim x · ln |x|.
x→0