## Precalculus (6th Edition) Blitzer

The function f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{2}}-9}{x-3}\text{ if }x\ne 3 \\ & \text{6 if }x=3 \end{align} \right. is not discontinuous for any number.
Consider the function f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{2}}-9}{x-3}\text{ if }x\ne 3 \\ & \text{6 if }x=3 \end{align} \right., Find the value of $f\left( x \right)$ at $x=3$, From the definition of the function, $f\left( 3 \right)=6$ The function is defined at the point $x=3$. Now find the value of $\,\underset{x\to 3}{\mathop{\lim }}\,f\left( x \right)$, \begin{align} & \,\,\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}-9}{x-3}=\,\underset{x\to 3}{\mathop{\lim }}\,\frac{\left( x-3 \right)\left( x+3 \right)}{\left( x-3 \right)}\, \\ & =\,\underset{x\to 3}{\mathop{\lim }}\,x+3 \\ & =3+3 \\ & =6 \end{align} Thus, $\,\,\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}-9}{x-3}=6$ From the above two steps, $\,\,\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}-9}{x-3}=6=f\left( 3 \right)$ Thus, the function satisfies all the properties of being continuous. Thus, the function is continuous at $x=3$. Hence, the function f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{2}}-9}{x-3}\text{ if }x\ne 3 \\ & \text{6 if }x=3 \end{align} \right. is not discontinuous for any number.