Page 142 - Calculus Demystified

CHAPTER 5
EXAMPLE 5.8 Indeterminate Forms 129
Evaluate the limit
x
lim x .
x→0 +
SOLUTION
x
We study the limit of f(x) = x by considering ln f(x) = x · ln x. We
rewrite this as
ln x
lim ln f(x) = lim .
x→0 + x→0 1/x
+
Both numerator and denominator tend to ±∞, so the quotient is indeterminate
of the form −∞/∞. Thus l’Hôpital’s Rule applies. The limit equals
1/x
lim = lim −x = 0.
x→0 −1/x 2 x→0 +
+
x
Now the only way that ln f(x) can tend to zero is if f(x) = x tends to 1. We
conclude that
x
lim x = 1.
x→0 +
EXAMPLE 5.9
Evaluate the limit
2 ln |x|
lim (1 + x ) .
x→0
SOLUTION
2
2 ln |x|
Let f(x) = (1 + x ) and consider ln f(x) = ln |x|· ln(1 + x ). This
expression is indeterminate of the form −∞ · 0.
We rewrite it as
2
ln(1 + x )
lim ,
x→0 1/ ln |x|
so that both the numerator and denominator tend to 0. So l’Hôpital’s Rule
applies and we have

2
2
2
2x/(1 + x ) 2x ln (|x|)
lim ln f(x) = lim = lim − .
2
2
x→0 x→0 −1/[x ln (|x|)] x→0 (1 + x )
The numerator tends to 0 (see Example 5.3) and the denominator tends to 1.
Thus
lim ln f(x) = 0.
x→0