# 009A Sample Final 2, Problem 2

Let

$f(x)=\left\{{\begin{array}{lr}{\frac {x^{2}-2x-3}{x-3}}&{\text{if }}x\neq 3\\5&{\text{if }}x=3\end{array}}\right.$ For what values of  $x$ is  $f$ continuous?

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

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