# 022 Exam 1 Sample A, Problem 3

Problem 3. Given a function $g(x)={\frac {x+5}{x^{2}-25}}$ ,

(a) Find the intervals where $g(x)$ is continuous.
(b). Find $\lim _{x\rightarrow 5}g(x)$ .
Foundations:
A function $f$ is continuous at a point $x_{0}$ if
$\lim _{x\rightarrow x_{_{0}}}f(x)=f\left(x_{0}\right).$ This can be viewed as saying the left and right hand limits exist, and are equal to the value of $f$ at $x_{0}$ .

Solution:

(a):
Note that
$g(x)\,\,=\,\,{\frac {x+5}{x^{2}-5}}\,\,=\,\,{\frac {x+5}{(x-5)(x+5)}}.$ In order to be continuous at a point $x_{0}$ , $f(x_{0})$ must exist. However, attempting to plug in $x=\pm 5$ results in division by zero. Therefore, in interval notation, we have that $f$ is continuous on
$(-\infty ,-5)\cup (-5,5)\cup (5,\infty ).$ (b):
Note that in order for the limit to exist, the limit from both the left and the right must be equal. But
${\begin{array}{rcl}\lim _{x\rightarrow 5^{+}}g(x)}&=&\lim _{x\rightarrow 5^{+}}{\frac {x+5}{(x-5)(x+5)}}}\\\\&=&\lim _{x\rightarrow 5^{+}}{\frac {1}{x-5}}}\\\\&=&{\frac {1}{\,\,0^{+}}}\rightarrow +\infty ,}\end{array}}$ while
${\begin{array}{rcl}\lim _{x\rightarrow 5^{-}}g(x)}&=&\lim _{x\rightarrow 5^{-}}{\frac {x+5}{(x-5)(x+5)}}}\\\\&=&\lim _{x\rightarrow 5^{-}}{\frac {1}{x-5}}}\\\\&=&{\frac {1}{\,\,0^{-}}}\rightarrow -\infty ,}\end{array}}$ where  $0^{-}$ can be thought of as "really small negative numbers approaching zero." Since the handed limits do not agree, the limit as $x$ approaches 5 does not exist.
(a): $f$ is continuous on $(-\infty ,-5)\cup (-5,5)\cup (5,\infty ).$ (b): $\lim _{x\rightarrow 5}g(x)$ does not exist.