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

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

Limits and Continuity 105

6.7 Solutions to Practice Problems

Part A The use of a calculator is not

allowed.

1. Using the product rule,
   limx→ 0 (x−5)(cosx)=
   [
   limx→ 0 (x−5)

][

limx→ 0 (cosx)

]

=(0−5)(cos 0)=(−5)(1)=−5.

(Note that cos 0=1.)

2. Rewrite lim
   x→b

x^3 −b^3

x^6 −b^6

as

lim

x→b

x^3 −b^3

(x^3 −b^3 )(x^3 +b^3 )

=lim

x→b

1

x^3 +b^3

.

Substitutex=band obtain

1

b^3 +b^3

=

1

2 b^3

.

3. Substitutingx=0 into the expression
   2 −

√

4 −x

x

leads to 0/0, which is an

indeterminate form. Thus, multiply both

the numerator and denominator by the

conjugate

(

2 +

√

4 −x

)

and obtain

limx→ 0

2 −

√

4 −x

x

(

2 +

√

4 −x

2 +

√

4 −x

)

=limx→ 0

4 −(4−x)

x

(

2 +

√

4 −x

)

=limx→ 0

x

x

(

2 +

√

4 −x

)

=limx→ 0

1

(

2 +

√

4 −x

)

=

1

(

2 +

√

4 −(0)

)=

1

4

.

4. Since the degree of the polynomial in the
   numerator is the same as the degree of the
   polynomial in the denominator,

xlim→∞

5 − 6 x

2 x+ 11

=−

6

2

=−3.

5. Since the degree of the polynomial in the
   numerator is 2 and the degree of the

polynomial in the denominator is 3,

lim

x→−∞

x^2 + 2 x− 3

x^3 + 2 x^2

=0.

6. The degree of the monomial in the
   numerator is 2, and the degree of the
   binomial in the denominator is 1. Thus,

xlim→∞

3 x^2

5 x+ 8

=∞.

7. Divide every term in both the numerator
   and denominator by the highest power of
   x. In this case, it isx. Thus, you have

lim

x→−∞

3 x

√x

x^2 − 4

x

.Asx→−∞,x=−

√

x^2.

Since the denominator involves a radical,

rewrite the expression as

x→lim−∞

3 x

√x

x^2 − 4

−

√

x^2

=xlim→−∞

3

−

√

1 −

4

x^2

=

3

−

√

1 − 0

=− 3.

8. limx→ 1 +f(x)=limx→ 1 +(x^2 ex)=eand
   limx→ 1 −f(x)=limx→ 1 −(ex)=e. Thus, limx→ 1 f(x)=e.


9. limx→∞ex=∞and limx→∞(1−x^3 )=∞.
   However, asx→∞, the rate of increase
   ofexis much greater than the rate of
   decrease of (1−x^3 ). Thus,

xlim→∞

ex

1 −x^3

=−∞.

10. Divide both numerator and denominator

byxand obtain limx→ 0

sin 3x

x

sin 4x

x

. Now rewrite

the limit as limx→ 0

3

sin 3x

3 x

4

sin 4x

4 x

=

3

4

limx→ 0

sin 3x

3 x

sin 4x

4 x

.

Asxapproaches 0, so do 3xand 4x.