Answer
The function $ f\left( x \right)=\left\{ \begin{align}
& 1-x\text{ if }x<1 \\
& 0\text{ if }x=1\text{ } \\
& {{x}^{2}}-1\text{ if }x>1
\end{align} \right.$ is continuous at $1$.
Work Step by Step
Consider the function $ f\left( x \right)=\left\{ \begin{align}
& 1-x\text{ if }x<1 \\
& 0\text{ if }x=1\text{ } \\
& {{x}^{2}}-1\text{ if }x>1
\end{align} \right.$.
Find the value of $ f\left( x \right)$ at $ a=1$,
From the definition of the function,
$ f\left( 1 \right)=0$
The function is defined at the point $ a=1$.
Now find the value of $\,\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$.
First find the left hand limit of $\,f\left( x \right)$.
The function $ f\left( x \right)=1-x $ when $ x<1$.
$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=1-1=0$
Now find the right hand limit of $\,f\left( x \right)$.
The function $ f\left( x \right)={{x}^{2}}-1$ when $ x>1$.
$\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{1}^{2}}-1=0$
Since the left hand limit and right hand limit are equal, that is $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$.
Thus, $\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)=0$.
From the above steps, $\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)=0=f\left( 1 \right)$.
Thus, the function satisfies all the properties of being continuous.
Hence, the function $ f\left( x \right)=\left\{ \begin{align}
& 1-x\text{ if }x<1 \\
& 0\text{ if }x=1\text{ } \\
& {{x}^{2}}-1\text{ if }x>1
\end{align} \right.$ is continuous at $1$.
