Answer
The function $ f\left( x \right)=\left\{ \begin{align}
& \frac{{{x}^{2}}-9}{x-3}\text{ if }x\ne 3 \\
& \text{6 if }x=3
\end{align} \right.$ is not discontinuous for any number.
Work Step by Step
Consider the function $ f\left( x \right)=\left\{ \begin{align}
& \frac{{{x}^{2}}-9}{x-3}\text{ if }x\ne 3 \\
& \text{6 if }x=3
\end{align} \right.$,
Find the value of $ f\left( x \right)$ at $ x=3$,
From the definition of the function,
$ f\left( 3 \right)=6$
The function is defined at the point $ x=3$.
Now find the value of $\,\underset{x\to 3}{\mathop{\lim }}\,f\left( x \right)$,
$\begin{align}
& \,\,\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}-9}{x-3}=\,\underset{x\to 3}{\mathop{\lim }}\,\frac{\left( x-3 \right)\left( x+3 \right)}{\left( x-3 \right)}\, \\
& =\,\underset{x\to 3}{\mathop{\lim }}\,x+3 \\
& =3+3 \\
& =6
\end{align}$
Thus, $\,\,\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}-9}{x-3}=6$
From the above two steps, $\,\,\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}-9}{x-3}=6=f\left( 3 \right)$
Thus, the function satisfies all the properties of being continuous.
Thus, the function is continuous at $ x=3$.
Hence, the function $ f\left( x \right)=\left\{ \begin{align}
& \frac{{{x}^{2}}-9}{x-3}\text{ if }x\ne 3 \\
& \text{6 if }x=3
\end{align} \right.$ is not discontinuous for any number.
