Answer
The statement “For the piecewise function defined by $ f\left( x \right)=\left\{ \begin{align}
& \frac{{{x}^{2}}-25}{x-5}\text{ if }x\ne 5 \\
& \text{12 if }x=5\text{ }
\end{align} \right.$, $\,\underset{x\to 5}{\mathop{\lim }}\,f\left( x \right)$ exists and is equal to $10$, so f is continuous at $5$” is false.
Work Step by Step
Consider the function, $ f\left( x \right)=\left\{ \begin{align}
& \frac{{{x}^{2}}-25}{x-5}\text{ if }x\ne 5 \\
& \text{12 if }x=5\text{ }
\end{align} \right.$
For a function to be continuous at a point a, the function must satisfy the following three conditions:
(a) f is defined at a.
(b) $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ exists.
(c) $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$
From the definition of the function,
$ f\left( 5 \right)=12$
The function is defined at $5$.
Thus, $\,\underset{x\to 5}{\mathop{\lim }}\,f\left( x \right)$ exists and is equal to $10$.
Thus, $\,\underset{x\to 5}{\mathop{\lim }}\,f\left( x \right)=10$
From the above two steps,
$\,\underset{x\to 5}{\mathop{\lim }}\,f\left( x \right)\ne f\left( 5 \right)$
Thus, the function do not satisfy the third condition of being continuous.
Therefore, the function f is not continuous at $5$.
Hence, the statement is false.
