## Precalculus (6th Edition) Blitzer

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.
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.