MA 3972-MA-Book May 7, 2018 9:52102 STEP 4. Review the Knowledge You Need to Score High
6.4 Rapid Review
- Findf(2) and limx→ 2 f(x) and determine iffis continuous atx=2. (See Figure 6.4-1.)
Answer: f(2)=2, lim
x→ 2
f(x)=4, and fis discontinuous atx=2.2024(4, 2)yxf(x)(2, 2)Figure 6.4-1- Evaluate limx→a
x^2 −a^2
x−a
.
Answer: xlim→a
(x+a)(x−a)
x−a
= 2 a.- Evaluate limx→∞
1 − 3 x^2
x^2 + 100 x+ 99
.
Answer: The limit is−3 since the polynomials in the numerator and denominator have
the same degree.- Determine iff(x)=
{
x+6 forx< 3
x^2 forx≥ 3
is continuous atx=3.Answer: The functionfis continuous sincef(3)=9, limx→ 3 +f(x)=limx→ 3 − f(x)=9, and
f(3)=limx→ 3 f(x).- Iff(x)=
{
ex forx= 0
5 forx= 0, find limx→ 0 f(x).Answer: limx→ 0 f(x)=1 since limx→ 0 + f(x)=xlim→ 0 −f(x)=1.- Evaluate limx→ 0
sin 6x
sin 2x
.
Answer: The limit is6
2
=3 since limx→ 0
sinx
x=1.
- Evaluate lim
x→ 5 −
x^2
x^2 − 25.
Answer: The limit is−∞since (x^2 −25) approaches 0 through negative values.