## Precalculus (6th Edition) Blitzer

No, the function f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{2}}-36}{x-6}\text{ if }x\ne 6 \\ & \text{13 if }x=6 \end{align} \right. is not continuous at $6$.
Consider the function f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{2}}-36}{x-6}\text{ if }x\ne 6 \\ & \text{13 if }x=6 \end{align} \right., First check whether the function is defined at the point $a$ or not. Find the value of $f\left( x \right)$ at $a=6$, Here use the second line of the piecewise function where $x=6$ because the first line of the piecewise function is not defined at 6. $f\left( 6 \right)=13$ The function is defined at the point $a=2$. Now find the value of $\,\underset{x\to 6}{\mathop{\lim }}\,\frac{{{x}^{2}}-36}{x-6}$, \begin{align} & \underset{x\to 6}{\mathop{\lim }}\,f\left( x \right)=\,\underset{x\to 6}{\mathop{\lim }}\,\frac{{{x}^{2}}-36}{x-6} \\ & =\,\underset{x\to 6}{\mathop{\lim }}\,\frac{\left( x-6 \right)\left( x+6 \right)}{\left( x-6 \right)}\, \\ & =\,\underset{x\to 6}{\mathop{\lim }}\,x+6 \\ & =6+6 \end{align} Further solve the above expression, $\underset{x\to 6}{\mathop{\lim }}\,f\left( x \right)=12$ Thus, $\,\underset{x\to 6}{\mathop{\lim }}\,\frac{{{x}^{2}}-36}{x-6}=12$ Now check whether $\,\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ or not. From the above, $\,\underset{x\to 6}{\mathop{\lim }}\,\frac{{{x}^{2}}-36}{x-6}=12\text{ and }f\left( 6 \right)=13$ Therefore, $\,\underset{x\to 6}{\mathop{\lim }}\,f\left( x \right)\ne f\left( 6 \right)$ Thus, the function does not satisfy the third property of being continuous. Hence, the function f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{2}}-36}{x-6}\text{ if }x\ne 6 \\ & \text{13 if }x=6 \end{align} \right. is not continuous at $6$.