Answer
The function $ f\left( x \right)=\left\{ \begin{align}
& x-5\text{ if }x\le 0 \\
& {{x}^{2}}+x-5\text{ if }x>0
\end{align} \right.$ is continuous at $0$.
Work Step by Step
Consider the function $ f\left( x \right)=\left\{ \begin{align}
& x-5\text{ if }x\le 0 \\
& {{x}^{2}}+x-5\text{ if }x>0
\end{align} \right.$,
Find the value of $ f\left( x \right)$ at $ a=0$,
$\begin{align}
& f\left( 0 \right)=0-5 \\
& =-5
\end{align}$
The function is defined at the point $ a=0$.
Now find the value of $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$.
First, find the left hand limit of $\,f\left( x \right)$.
The function $ f\left( x \right)=x-5$ when $ x<0$.
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0-5=-5$
Now find the right hand limit of $\,f\left( x \right)$.
The function $ f\left( x \right)={{x}^{2}}+x-5$ when $ x>0$.
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{0}^{2}}+0-5=-5$
Since the left hand limit and right hand limit are equal, that is $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=-5=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$
Thus, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=-5$
From the above steps, $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=-5=f\left( 0 \right)$
Thus, the function satisfies all the properties of being continuous.
Hence, the function $ f\left( x \right)=\left\{ \begin{align}
& x-5\text{ if }x\le 0 \\
& {{x}^{2}}+x-5\text{ if }x>0
\end{align} \right.$ is continuous at $0$.
