Answer
The function $ f\left( x \right)=\left\{ \begin{align}
& \frac{{{x}^{2}}-1}{x-1}\text{ if }x\ne 1 \\
& 2\text{ if }x=1
\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}}-1}{x-1}\text{ if }x\ne 1 \\
& 2\text{ if }x=1
\end{align} \right.$,
Find the value of $ f\left( x \right)$ at $ x=1$,
From the definition of the function,
$ f\left( 1 \right)=2$
The function is defined at the point $ x=1$.
Now find the value of $\,\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$,
$\begin{align}
& \,\,\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{2}}-1}{x-1}=\,\underset{x\to 1}{\mathop{\lim }}\,\frac{\left( x-1 \right)\left( x+1 \right)}{\left( x-1 \right)}\, \\
& =\,\underset{x\to 1}{\mathop{\lim }}\,x+1 \\
& =1+1 \\
& =2
\end{align}$
Thus, $\,\,\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{2}}-1}{x-1}=\,2$
From the above two steps, $\,\,\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{2}}-1}{x-1}=\,2=f\left( 1 \right)$
Thus, the function satisfies all the properties of being continuous.
Thus, the function is continuous at $ x=1$.
Hence, the function $ f\left( x \right)=\left\{ \begin{align}
& x-2\text{ if }x\le 2 \\
& {{x}^{2}}-1\text{ if }x>2
\end{align} \right.$ is not discontinuous for any number.
