Page 137 - Calculus Demystified

CHAPTER 5
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5.1.2 L’HÔPITAL’S RULE Indeterminate Forms
Theorem 5.1 (l’Hôpital’s Rule)
Let f(x) and g(x) be differentiable functions on (a, c) ∪ (c, b). If
lim f(x) = lim g(x) = 0
x→c x→c
then

f(x) f (x)
lim = lim ,
x→c g(x) x→c g (x)

provided this last limit exists as a finite or infinite limit.
Let us learn how to use this new result.
EXAMPLE 5.1
Evaluate
ln x
lim .
2
x→1 x + x − 2
SOLUTION
We first notice that both the numerator and denominator have limit zero
as x tends to 1. Thus the quotient is indeterminate at 1 and of the form 0/0.
l’Hôpital’s Rule therefore applies and the limit equals

(d/dx)(ln x)
lim ,
2
x→1 (d/dx)(x + x − 2)
provided this last limit exists. The last limit is
1/x 1
lim = lim .
2
x→1 2x + 1 x→1 2x + x
Therefore we see that
ln x 1
lim = .
2
x→1 x + x − 2 3
2
You Try It: Apply l’Hôpital’s Rule to the limit lim x→2 sin(πx)/(x − 4).
You Try It: Usel’Hôpital’sRuletoevaluatelim h→0 (sin h/h)andlim h→0 (cos h−
1/h). These limits are important in the theory of calculus.
EXAMPLE 5.2
Evaluate the limit
x 3
lim .
x→0 x − sin x