Answer
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 discontinuous for $ x=2$.
Work Step by Step
Consider 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.$
Find the value of $ f\left( x \right)$ at $ x=2$,
From the definition of the function, for $ x=2$, $ f\left( x \right)=x-2$
Then the value of $ f\left( x \right)$ at $ x=2$ is,
$\begin{align}
& f\left( 2 \right)=2-2 \\
& =0
\end{align}$
The function is defined at the point $ x=2$.
Now find the value of $\,\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)$,
First find the left-hand limit of $\,f\left( x \right)$,
That is,
$\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=2-2=0$
Now find the right-hand limit of $\,f\left( x \right)$,
That is,
$\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{2}^{2}}-1=3$
Since, the left-hand limit and right-hand limit are not equal, that is $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)\ne \underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$
Thus, $\,\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)$ does not exist.
Thus, the function does not satisfy the second property of being continuous.
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 discontinuous at $ x=2$.
