007A Sample Final 2, Problem 2 Detailed Solution

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?

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Background Information:
$f(x)$  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  $(-\infty ,3)\cup (3,\infty ).$

Final Answer:
$f(x)$  is continuous on  $(-\infty ,3)\cup (3,\infty ).$

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