Answer
Yes, the function $ f\left( x \right)=\frac{{{x}^{2}}+4}{x-2}$ is continuous at $3$.
Work Step by Step
Consider the function $ f\left( x \right)=\frac{{{x}^{2}}+4}{x-2}$,
First check whether the function is defined at the point $ a $ or not.
Find the value of $ f\left( x \right)$ at $ a=3$,
$\begin{align}
& f\left( 3 \right)=\frac{{{\left( 3 \right)}^{2}}+4}{\left( 3 \right)-2} \\
& =\frac{9+4}{3-2} \\
& =13
\end{align}$
The function is defined at the point $ a=3$.
Now find the value of $\,\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}+4}{x-2}$,
$\begin{align}
& \,\underset{x\to 3}{\mathop{\lim }}\,f\left( x \right)=\,\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}+4}{x-2} \\
& =\frac{\,\underset{x\to 3}{\mathop{\lim }}\,\left( {{x}^{2}}+4 \right)}{\,\underset{x\to 3}{\mathop{\lim }}\,\left( x-2 \right)}\, \\
& =\frac{{{\left( 3 \right)}^{2}}+4}{3-2} \\
& =13
\end{align}$
Thus, $\,\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}+4}{x-2}=13$
Now check whether $\,\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ or not.
From the above, $\,\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}+4}{x-2}=13\text{ and }f\left( 3 \right)=13$
Therefore, $\,\underset{x\to 3}{\mathop{\lim }}\,f\left( x \right)=f\left( 3 \right)$
Thus, the function satisfies all the properties of being continuous.
Hence, the function $ f\left( x \right)=\frac{{{x}^{2}}+4}{x-2}$ is continuous at $3$.
