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SOLUTION CHAPTER 5 Indeterminate Forms
This may be rewritten as
ln |x|
lim .
x→0 1/x 3
Notice that the numerator tends to −∞ and the denominator tends to ±∞ as
x → 0. Thus the quotient is indeterminate at 0 of the form −∞/ +∞. So we
may apply l’Hôpital’s Rule for infinite limits to see that the limit equals
1/x
3
lim = lim −x /3 = 0.
x→0 −3x −4 x→0
Yet another version of l’Hôpital’s Rule, this time for unbounded intervals, is
this: Let f and g be differentiable functions on an interval of the form [A, +∞).
If lim x→+∞ f(x) = lim x→+∞ g(x) = 0oriflim x→+∞ f(x) =±∞ and
lim x→+∞ g(x) =±∞, then

f(x) f (x)
lim = lim
x→+∞ g(x) x→+∞ g (x)

provided that this last limit exists either as a finite or infinite limit. The same result
holds for f and g defined on an interval of the form (−∞,B] and for the limit as
x →−∞.
EXAMPLE 5.4
Evaluate
x 4
lim .
x→+∞ e x
SOLUTION
We first notice that both the numerator and the denominator tend to +∞ as
x →+∞. Thus the quotient is indeterminate at +∞ of the form +∞/ +∞.
Therefore the new version of l’Hôpital’s Rule applies and our limit equals

4x 3
lim .
x→+∞ e x
Again the numerator and denominator tend to +∞ as x →+∞, so we once
more apply l’Hôpital’s Rule. The limit equals
12x 2
lim = 0.
x→+∞ e x
We must apply l’Hôpital’s Rule two more times. We first obtain
24x
lim
x→+∞ e x