5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book May 7, 2018 9:52

Limits and Continuity 91

[−3, 3] by [−3, 3]

Figure 6.1-8

Example 3

Find the limit if it exists: limy→ 0
y^2
1 −cosy

.

Substituting 0 in the expression would lead to 0/0. Multiplying both the numera-

tor and denominator by the conjugate (1+cosy) produces limy→ 0
y^2
1 −cosy

·

(1+cosy)

(1+cosy)

=limy→ 0
y^2 (1+cosy)
1 −cos^2 y

=limy→ 0

y^2 (1+cosy)

sin^2 y

=limy→ 0

y^2

sin^2 y

·limy→ 0 (1+cos^2 y)

=limy→ 0

(

y

siny

) 2

·limy→ 0 (1+cos^2 y)=

(

limy→ 0

y

siny

) 2

·limy→ 0 (1+cos^2 y)=(1)^2 (1+1)=2.

(Note that limy→ 0
y
siny
=limy→ 0

1

siny

y

=

limy→ 0 (1)

limy→ 0

siny

y

=

1

1

=1.) Verify your result with a calculator.

(See Figure 6.1-9.)

[−8, 8] by [−2, 10]
Figure 6.1-9
Example 4

Find the limit if it exists: limx→ 0
3 x
cosx

.

Using the quotient rule for limits, you have limx→ 0
3 x
cosx

=

limx→ 0 (3x)

limx→ 0 (cosx)

=

0

1

=0.

Verify your result with a calculator. (See Figure 6.1-10.)

[−10, 10] by [−30, 30]

Figure 6.1-10