5 Steps to a 5 AP Calculus BC 2019

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

1. Ify=ex^3 , find
   dy
   dx

.

Answer: Using the chain rule,

dy

dx

=

(

ex^3

)

(3x^2 ).

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

)

=−

1

2

.