# 022 Exam 1 Sample A, Problem 3

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

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

Solution:

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