Answer
The function $ f\left( x \right)=\left\{ \begin{align}
& x-1\text{ if }x\le 1 \\
& {{x}^{2}}\text{ if }x>1
\end{align} \right.$ is discontinuous for $ x=1$.
Work Step by Step
Consider the function $ f\left( x \right)=\left\{ \begin{align}
& x-1\text{ if }x\le 1 \\
& {{x}^{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, for $ x=1$, $ f\left( x \right)=x-1$
Then the value of $ f\left( x \right)$ at $ x=1$ is,
$\begin{align}
& f\left( 1 \right)=1-1 \\
& =0
\end{align}$
The function is defined at the point $ x=1$.
Now find the value of $\,\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$,
First find the left-hand limit of $\,f\left( x \right)$,
That is,
$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=1-1=0$
Now find the right-hand limit of $\,f\left( x \right)$,
That is,
$\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{1}^{2}}=1$
Since, the left-hand limit and right-hand limit are not equal, that is $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)\ne \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$.
Thus, $\,\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$ 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-1\text{ if }x\le 1 \\
& {{x}^{2}}\text{ if }x>1
\end{align} \right.$ is discontinuous at $ x=1$.
