Page 143 - Calculus Demystified

Indeterminate Forms
CHAPTER 5
130
But the only way that ln f(x) can tend to zero is if f(x) tends to 1. We conclude
that
2 ln |x|
lim (1 + x ) = 1.
x→0
x
You Try It: Evaluate the limit lim x→0 +(1/x) .
You Try It: Evaluate the limit lim x→0 +(1 + x) 1/x . In fact this limit gives an
important way to define Euler’s constant e (see Sections 1.9 and 6.2.3).

5.2.4 PUTTING TERMS OVER A COMMON

DENOMINATOR
Many times a simple algebraic manipulation involving fractions will put a limit
into a form which can be studied using l’Hôpital’s Rule.

EXAMPLE 5.10
Evaluate the limit

1 1
lim − .
x→0 sin 3x 3x
SOLUTION
We put the fractions over a common denominator to rewrite our limit as

3x − sin 3x
lim .
x→0 3x · sin 3x
Both numerator and denominator vanish as x → 0. Thus the quotient has
indeterminate form 0/0. By l’Hôpital’s Rule, the limit is therefore equal to
3 − 3 cos 3x
lim .
x→0 3 sin 3x + 9x cos 3x
This quotient is still indeterminate; we apply l’Hôpital’s Rule again to obtain

9 sin 3x
lim = 0.
x→0 18 cos 3x − 27x sin 3x
EXAMPLE 5.11
Evaluate the limit

1 1
lim − .
x→0 4x e 4x − 1