Answer
No, the function $ f\left( x \right)=\left\{ \begin{align}
& \frac{{{x}^{2}}-4}{x-2}\text{ if }x\ne 2 \\
& 5\text{ if }x=2
\end{align} \right.$ is not continuous at $2$.
Work Step by Step
Consider the function $ f\left( x \right)=\left\{ \begin{align}
& \frac{{{x}^{2}}-4}{x-2}\text{ if }x\ne 2 \\
& 5\text{ if }x=2
\end{align} \right.$,
First check whether the function is defined at the point $ a $ or not.
Find the value of $ f\left( x \right)$ at $ a=2$,
Here use the second line of the piecewise function where $ x=2$ because the first line of the piecewise function is not defined at 2
$ f\left( 2 \right)=5$
The function is defined at the point $ a=2$.
Now find the value of $\,\underset{x\to 2}{\mathop{\lim }}\,\frac{{{x}^{2}}-4}{x-2}$,
$\begin{align}
& \underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 2}{\mathop{\lim }}\,\frac{{{x}^{2}}-4}{x-2} \\
& =\,\underset{x\to 2}{\mathop{\lim }}\,\frac{\left( x-2 \right)\left( x+2 \right)}{\left( x-2 \right)}\, \\
& =\,\underset{x\to 2}{\mathop{\lim }}\,\left( x+2 \right) \\
& =2+2
\end{align}$
Further solve the above expression,
$\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)=4$
Thus, $\,\underset{x\to 2}{\mathop{\lim }}\,\frac{{{x}^{2}}-4}{x-2}=4$
Now check whether $\,\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ or not.
From the above, $\,\underset{x\to 2}{\mathop{\lim }}\,\frac{{{x}^{2}}-4}{x-2}=4\text{ and }f\left( 2 \right)=5$
Therefore, $\,\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)\ne f\left( 2 \right)$
Thus, the function does not satisfy the third property of being continuous.
Hence, the function $ f\left( x \right)=\left\{ \begin{align}
& \frac{{{x}^{2}}-4}{x-2}\text{ if }x\ne 2 \\
& 5\text{ if }x=2
\end{align} \right.$ is not continuous at $2$.
