Answer
The function $f\left( x \right)=\left\{ \begin{align}
& \sqrt{3-x}\text{ if }x\le 3 \\
& {{x}^{2}}-3x\text{ if }x>3
\end{align} \right.$ is continuous at $3$.
Work Step by Step
Consider the function $f\left( x \right)=\left\{ \begin{align}
& \sqrt{3-x}\text{ if }x\le 3 \\
& {{x}^{2}}-3x\text{ if }x>3
\end{align} \right.$ ,
Find the value of $f\left( x \right)$ at $a=3$ ,
From the definition of the function, for $a=3$, the function takes the value $\sqrt{3-x}$ ,
Thus,
$\begin{align}
& f\left( 3 \right)=\sqrt{3-3} \\
& =0
\end{align}$
The function is defined at the point $a=3$.
Now find the value of $\,\underset{x\to 3}{\mathop{\lim }}\,f\left( x \right)$ ,
First find the left hand limit of $\,f\left( x \right)$ ,
As x nears $3$ from the left, the function $\,f\left( x \right)$ takes the value $\sqrt{3-x}$ ,
Thus, $\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\sqrt{3-3}=0$
Now find the right hand limit of $\,f\left( x \right)$ ,
As x nears $3$ from the right, the function $\,f\left( x \right)$ takes the value ${{x}^{2}}-3x$ ,
Thus, $\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{3}^{2}}-3\left( 3 \right)=0$
Since the left hand limit and right hand limit are equal, that is $\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0=\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)$
Thus, $\underset{x\to 3}{\mathop{\lim }}\,f\left( x \right)=0$
From the above steps, $\underset{x\to 3}{\mathop{\lim }}\,f\left( x \right)=0=f\left( 3 \right)$
Since, the function satisfies all the properties of being continuous,
Hence, the function $f\left( x \right)=\left\{ \begin{align}
& \sqrt{3-x}\text{ if }x\le 3 \\
& {{x}^{2}}-3x\text{ if }x>3
\end{align} \right.$ is continuous at $3$.
