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EXAMPLE 2.2 The Indeterminate Form ∞
30 ∞
e x
Evaluate lim .
x→∞ x
20 ∞
Solution This has the form and from the graph in Figure 4.15, it appears that
∞
the function grows larger and larger, without bound, as x → ∞. Applying
l’Ĥopital’s Rule confirms our suspicions, as
10
d (e )
x
e x dx e x
lim = lim = lim =∞.
x x→∞ x x→∞ d x→∞ 1
1 2 3 4 5 (x)
dx
FIGURE 4.15
e x For some limits, you may need to apply l’Ĥopital’s Rule repeatedly. Just be careful
y = to verify the hypotheses at each step.
x

y EXAMPLE 2.3 A Limit Requiring Two Applications of L’H ̂ opital’s
Rule
0.6 x 2
Evaluate lim .
x→∞ e x
0.4 Solution First, note that this limit has the form ∞ . From the graph in Figure 4.16, it
∞
seems that the function tends to 0 as x → ∞. Applying l’Ĥopital’s Rule twice, we get
d 2
0.2 x 2 dx (x ) 2x ( ∞ )
lim = lim = lim
x→∞ e x x→∞ d (e ) x→∞ e x ∞
x
x dx
2 4 6 8 10 d (2x)
FIGURE 4.16 = lim dx = lim 2 = 0,
x
y = x 2 x→∞ d (e ) x→∞ e x
e x dx
as expected.
REMARK 2.2

A very common error is to apply l’Ĥopital’s Rule indiscriminately, without first
checking that the limit has the indeterminate form 0 or ∞ . Students also
∞
0
sometimes incorrectly compute the derivative of the quotient, rather than the
quotient of the derivatives. Be very careful here.

EXAMPLE 2.4 An Erroneous Use of L’H ̂ opital’s Rule
y
Find the mistake in the string of equalities

lim x 2 = lim 2x = lim 2 = 2 = 2. This is incorrect !
x
x→0 e − 1 x→0 e x x→0 e x 1
x
3 Solution From the graph in Figure 4.17, we can see that the limit is approximately 0,
x 2
0
so 2 appears to be incorrect. The first limit, lim x , has the form and the
Copyright © McGraw-Hill Education FIGURE 4.17 fore, the first equality, lim x lim = lim = lim 2x = 0 = 0. x→0 e x = 0
x→0 e − 1
0
2
x
functions f(x) = x and g(x) = e − 1 satisfy the hypotheses of l’Ĥopital’s Rule. There-
-3
2
2x
x
2x
, holds. However, notice that lim
x→0 e − 1
1
x
x→0 e
= 0 and l’Ĥopital’s Rule does not apply here. The correct evaluation is then
2
x
x
2
y =
x→0 e
x
x→0 e − 1
1
e − 1
x
x
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