Answer
Yes, the function $ f\left( x \right)=\frac{x-7}{x+7}$ is continuous at $7$.
Work Step by Step
Consider the function $ f\left( x \right)=\frac{x-7}{x+7}$,
First check whether the function is defined at the point $ a $ or not.
Find the value of $ f\left( x \right)$ at $ a=7$,
$\begin{align}
& f\left( 7 \right)=\frac{7-7}{7+7} \\
& =0
\end{align}$
The function is defined at the point $ a=7$.
Now find the value of $\,\underset{x\to 7}{\mathop{\lim }}\,\frac{x-7}{x+7}$,
$\begin{align}
& \underset{x\to 7}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 7}{\mathop{\lim }}\,\frac{x-7}{x+7} \\
& =\frac{\,\underset{x\to 7}{\mathop{\lim }}\,\left( x-7 \right)}{\,\underset{x\to 7}{\mathop{\lim }}\,\left( x+7 \right)}\, \\
& =\,\frac{7-7}{7+7}
\end{align}$
Thus, $\,\underset{x\to 7}{\mathop{\lim }}\,\frac{x-7}{x+7}=0$
Now check whether $\,\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ or not.
From the above two steps, $\,\underset{x\to 7}{\mathop{\lim }}\,\frac{x-7}{x+7}=0\text{ and }f\left( 7 \right)=0$
Therefore, $\,\underset{x\to 7}{\mathop{\lim }}\,f\left( x \right)=f\left( 7 \right)$
Thus, the function satisfies all the properties of being continuous.
Hence, the function $ f\left( x \right)=\frac{x-7}{x+7}$ is continuous at $7$.
