# 009A Sample Final 2, Problem 2

Let

${\displaystyle f(x)=\left\{{\begin{array}{lr}{\frac {x^{2}-2x-3}{x-3}}&{\text{if }}x\neq 3\\5&{\text{if }}x=3\end{array}}\right.}$

For what values of  ${\displaystyle x}$  is  ${\displaystyle f}$  continuous?

Foundations:
${\displaystyle f(x)}$  is continuous at  ${\displaystyle x=a}$  if
${\displaystyle \lim _{x\rightarrow a^{+}}f(x)=\lim _{x\rightarrow a^{-}}f(x)=f(a).}$

Solution:

Step 1:
Based on the description of  ${\displaystyle f(x),}$
we know  ${\displaystyle f(x)}$  is continuous on
${\displaystyle (-\infty ,3)\cup (3,\infty ).}$
Now, we need to see if  ${\displaystyle f(x)}$  is continuous at  ${\displaystyle x=3.}$
Step 2:
We have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 3^{+}}f(x)}&=&\displaystyle {\lim _{x\rightarrow 3^{+}}{\frac {x^{2}-2x-3}{x-3}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 3^{+}}{\frac {(x-3)(x+1)}{x-3}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 3^{+}}x+1}\\&&\\&=&\displaystyle {4.}\end{array}}}$
Similarly,
${\displaystyle \lim _{x\rightarrow 3^{-}}f(x)=4.}$
Since
${\displaystyle \lim _{x\rightarrow 3^{-}}f(x)=\lim _{x\rightarrow 3^{+}}f(x)}$
we have
${\displaystyle \lim _{x\rightarrow 3}f(x)=4.}$
Step 3:
But, since
${\displaystyle f(3)=5,}$
we have
${\displaystyle \lim _{x\rightarrow 3}f(x)\neq f(3).}$
Therefore,  ${\displaystyle f(x)}$  is not continuous at  ${\displaystyle x=3.}$
${\displaystyle f(x)}$  is continuous only on  ${\displaystyle (-\infty ,3)\cup (3,\infty ).}$

${\displaystyle f(x)}$  is continuous on  ${\displaystyle (-\infty ,3)\cup (3,\infty ).}$