MA 3972-MA-Book May 7, 2018 9:52
Limits and Continuity 93[−5, 7] by [−40, 20]
Figure 6.2-1Example 2
Find: limx→ 3 −
x^2
x^2 − 9
.
Factor the denominator obtaining limx→ 3 −
x^2
x^2 − 9
=limx→ 3 −
x^2
(x−3)(x+3)
. The limit of the numer-
ator is 9 and the limit of the denominator is (0)(6)=0 through negative values. Therefore,
limx→ 3 −
x^2
x^2 − 9
=−∞. Verify your result with a calculator. (See Figure 6.2-2.)
[−10, 10] by [−10, 10]
Figure 6.2-2Example 3
Find: lim
x→ 5 −
√
25 −x^2
x− 5.
Substituting 5 into the expression leads to 0/0. Factor the numerator
√
√^25 −x^2 into
(5−x)(5+x). Asx → 5 −,(x−5) < 0. Rewrite (x−5) as−(5−x). Asx → 5 −,
(5−x) > 0 and thus, you may express (5 − x)as
√
(5−x)^2 =√
(5−x)(5−x).Therefore, (x−5)=−(5−x)=−
√
(5−x)(5−x). Substituting these equivalent expres-sions into the original problem, you have limx→ 5 −
√
25 −x^2
x− 5
= limx→ 5 −√
(5−x)(5+x)
√
(5−x)(5−x)=
−limx→ 5 −
√
(5−x)(5+x)
(5−x)(5−x)
=−limx→ 5 −√
(5+x)
(5−x). The limit of the numerator is 10 and the limit
of the denominator is 0 through positive values. Thus, the lim
x→ 5 −
√
25 −x^2
x− 5