Answer
Yes, the function $ f\left( x \right)=\frac{{{x}^{2}}+6}{x-5}$ is continuous at $6$.
Work Step by Step
Consider the function $ f\left( x \right)=\frac{{{x}^{2}}+6}{x-5}$,
First check whether the function is defined at the point $ a $ or not.
Find the value of $ f\left( x \right)$ at $ a=6$,
$\begin{align}
& f\left( 6 \right)=\frac{{{\left( 6 \right)}^{2}}+6}{\left( 6 \right)-5} \\
& =\frac{36+6}{6-5} \\
& =42
\end{align}$
The function is defined at the point $ a=6$.
Now find the value of $\,\underset{x\to 6}{\mathop{\lim }}\,\frac{{{x}^{2}}+6}{x-5}$,
$\begin{align}
& \,\underset{x\to 6}{\mathop{\lim }}\,f\left( x \right)=\,\underset{x\to 6}{\mathop{\lim }}\,\frac{{{x}^{2}}+6}{x-5} \\
& =\frac{\,\underset{x\to 6}{\mathop{\lim }}\,\left( {{x}^{2}}+6 \right)}{\,\underset{x\to 6}{\mathop{\lim }}\,\left( x-5 \right)}\, \\
& =\frac{{{\left( 6 \right)}^{2}}+6}{6-5} \\
& =42
\end{align}$
Thus, $\,\underset{x\to 6}{\mathop{\lim }}\,\frac{{{x}^{2}}+6}{x-5}=42$
Now check whether $\,\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ or not.
From the above, $\,\underset{x\to 6}{\mathop{\lim }}\,\frac{{{x}^{2}}+6}{x-5}=42\text{ and }f\left( 6 \right)=42$
Therefore, $\,\underset{x\to 6}{\mathop{\lim }}\,f\left( x \right)=f\left( 6 \right)$
Thus, the function satisfies all the properties of being continuous.
Hence, the function $ f\left( x \right)=\frac{{{x}^{2}}+6}{x-5}$ is continuous at $6$.
