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

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

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

Example 2

Find the limit: limx→π 3 xsinx.

Using the product rule, limx→π 3 xsinx=

(

xlim→π^3 x

)(

xlim→πsinx

)

=(3π)(sinπ)=(3π)(0)=0.

Example 3

Find the limit: limt→ 2

t^2 − 3 t+ 2

t− 2

.

Factoring and simplifying: limt→ 2

t^2 − 3 t+ 2

t− 2

=limt→ 2

(t−1)(t−2)

(t−2)

=limt→ 2 (t−1)=(2−1)= 1.

(Note that had you substitutedt=2 directly in the original expression, you would have

obtained a zero in both the numerator and denominator.)

Example 4

Find the limit: limx→b

x^5 −b^5

x^10 −b^10

.

Factoring and simplifying: limx→b

x^5 −b^5

x^10 −b^10

=xlim→b

x^5 −b^5

(x^5 −b^5 )(x^5 +b^5 )

=xlim→b

1

x^5 +b^5

=

1

b^5 +b^5

=

1

2 b^5

.

Example 5

Find the limit: limt→ 0

√

t+ 2 −

√

2

t

.

Multiplying both the numerator and the denominator by the conjugate of the numerator,

(√

t+ 2 +

√

2

)

, yields limt→ 0

√

t+ 2 −

√

2

t

(√

t+ 2 +

√

2

√

t+ 2 +

√

2

)

=limt→ 0

t+ 2 − 2

t

(√

t+ 2 +

√

2

)

=limt→ 0

t

t

(√

t+ 2 +

√

2

)=limt→ 0

1

(√

t+ 2 +

√

2

)=

1

√

0 + 2 +

√

2

=

1

2

√

2

=

1

2

√

2

(√

2

√

2

)

=

√

2

4

.

(Note that substituting 0 directly into the original expression would have produced a 0 in

both the numerator and denominator.)