## Precalculus (6th Edition) Blitzer

The statement “For the function $f\left( x \right)=\frac{1}{x-3},f$ is not defined at $3$, so f is discontinuous at $3$” is true.
Consider the function $f\left( x \right)=\frac{1}{x-3}$, 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) $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ Since the function is not defined at $3$, so the function does not satisfy the first condition of being continuous. Therefore, the function f is discontinuous at $3$. Hence, the statement is true.