Differentiation 95
L’Hôpital’sRule for Indeterminate Forms
Letlimrepresent one of the limits: limx→c, limx→c+, limx→c−, limx→∞,orx→lim−∞. Suppose f(x) andg(x)
are differentiable, andg′(x)/=0 nearc, except possibly atc, and suppose limf(x)=0 and
limg(x)=0, then the lim
f(x)
g(x)
is an indeterminate form of the type
0
0
. Also, if limf(x)=
±∞and limg(x)=±∞, then the lim
f(x)
g(x)
is an indeterminate form of the type
∞
∞
.In
both cases,
0
0
and
∞
∞
,L’Hoˆpital’sRule states that lim
f(x)
g(x)
=lim
f′(x)
g′(x)
.
Example 4
Find limx→ 0
1 −cosx
x^2
, if it exists.
Since limx→ 0 (1−cosx)=0 and limx→ 0 (x^2 )=0, this limit is an inderminate form. Taking the
derivatives,
d
dx
(1−cosx)=sinxand
d
dx
(x^2 )= 2 x.ByL’Hoˆpital’sRule, limx→ 0
1 −cosx
x^2
=
limx→ 0
sinx
2 x
=
1
2
limx→ 0
sinx
x
=
1
2
.
Example 5
Find limx→∞x^3 e−x^2 , if it exists.
Rewriting limx→∞x^3 e−x^2 as limx→∞
(
x^3
ex^2
)
shows that the limit is an indeterminate form, since
xlim→∞(x^3 )=∞and limx→∞
(
ex^2
)
=∞. Differentiating and applyingL’Hoˆpital’sRule means
that limx→∞
(
x^3
ex^2
)
=xlim→∞
(
3 x^2
2 xex^2
)
=
3
2
xlim→∞
(x
ex^2
)
. Unfortunately, this new limit is also
indeterminate. However, it is possible to applyL’Hoˆpital’sRule again, so
3
2
xlim→∞
(x
ex^2
)
equals to
3
2
xlim→∞
(
1
2 xex^2
)
. This expression approaches zero asx becomes large, so
xlim→∞x^3 e−x^2 =0.
6.7 Rapid Review
- Ify=ex^3 , find
dy
dx
.
Answer: Using the chain rule,
dy
dx
=
(
ex^3
)
(3x^2 ).
- Evaluate limh→ 0
cos
(
π
6
+h
)
−cos
(
π
6
)
h
.
Answer: The limit is equivalent to
d
dx
cosx
∣∣
∣∣
x=π 6
=−sin
(
π
6
)
=−