Answer
Consider the piecewise function defined by $ f\left( x \right)=\left\{ \begin{align}
& 1-x\text{ if }x<1 \\
& \text{0 if }x=1\text{ } \\
& {{x}^{2}}-1\text{ if }x>1
\end{align} \right.$. We find $\,\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ using $ f\left( x \right)=$ $1-x $. We find $\,\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ using $ f\left( x \right)=$ ${{x}^{2}}-1$. The function’s definition indicates that $ f\left( 1 \right)=$ $0$. If we determine that $\,\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( 1 \right)$, we can conclude that the function is continuous at $1$.
Work Step by Step
Consider the function, $ f\left( x \right)=\left\{ \begin{align}
& 1-x\text{ if }x<1 \\
& \text{0 if }x=1\text{ } \\
& {{x}^{2}}-1\text{ if }x>1
\end{align} \right.$
The value $\,\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is the value of the function as x nears $1$ from the left.
As x nears $1$ from the left, the value of the function is $ f\left( x \right)=1-x $.
The value $\,\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is the value of the function as x nears $1$ from the right.
As x nears $1$ from the right, the value of the function is $ f\left( x \right)={{x}^{2}}-1$
From the definition of the function, the value of $ f\left( 1 \right)=0$
If $\,\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( 1 \right)$, then the function satisfies all the conditions of being continuous, so the function is continuous at $1$.
