Question

A function $f\left(x\right)$ is defined as $f\left(x\right)=\{\begin{array}{c}\frac{x^2-9}{x-3};ifx\ne 3\\ 6;ifx=3\end{array}$
Show that $f\left(x\right)$ is continuous at $x=3$.

Answer 1

Given,

$f\left(x\right)=\{\begin{array}{cc}\frac{x^2-9}{x-3},& \text{if}x\ne 3\\ 6,& \text{if}x=3\end{array}$

LHL at $=3$

$\underset{x\to 3^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(3-h)=\underset{h\to 0}{lim}\frac{{(3-h)}^2-9}{(3-h)-3}=\underset{h\to 0}{lim}\frac{h^2-6h}{-h}=\underset{h\to 0}{lim}(6-h)=6$

RHL at $=3$

$\underset{x\to 3^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(3+h)=\underset{h\to 0}{lim}\frac{{(3+h)}^2-9}{(3+h)-3}=\underset{h\to 0}{lim}\frac{h^2+6h}{h}=\underset{h\to 0}{lim}(6+h)=6$

Also,

$f\left(3\right)=6\left(\text{Given}\right)$

Now, since

$\underset{x\to 3^{-}}{lim}f\left(x\right)=\underset{x\to 3}{lim}f\left(x\right)=\underset{x\to 3^{+}}{lim}f\left(x\right)$

Hence $f\left(x\right)$ is continuous at $x=3$.