5 Steps to a 5 AP Calculus BC 2019

56 STEP 4. Review the Knowledge You Need to Score High

[–3,3] by [–3,3]

Figure 5.1-8

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 numerator 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 5.1-9.)

[–8,8] by [–2,10]

Figure 5.1-9

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

=

xlim→ 0 (3x)

xlim→ 0 (cosx)

=

0

1

=0. Verify your

result with a calculator. (See Figure 5.1-10.)

[–10,10] by [–30,30]

Figure 5.1-10