## Precalculus (6th Edition) Blitzer

No, the function f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{2}}-4}{x-2}\text{ if }x\ne 2 \\ & 5\text{ if }x=2 \end{align} \right. is not continuous at $2$.
Consider the function f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{2}}-4}{x-2}\text{ if }x\ne 2 \\ & 5\text{ if }x=2 \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 $a=2$, Here use the second line of the piecewise function where $x=2$ because the first line of the piecewise function is not defined at 2 $f\left( 2 \right)=5$ The function is defined at the point $a=2$. Now find the value of $\,\underset{x\to 2}{\mathop{\lim }}\,\frac{{{x}^{2}}-4}{x-2}$, \begin{align} & \underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 2}{\mathop{\lim }}\,\frac{{{x}^{2}}-4}{x-2} \\ & =\,\underset{x\to 2}{\mathop{\lim }}\,\frac{\left( x-2 \right)\left( x+2 \right)}{\left( x-2 \right)}\, \\ & =\,\underset{x\to 2}{\mathop{\lim }}\,\left( x+2 \right) \\ & =2+2 \end{align} Further solve the above expression, $\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)=4$ Thus, $\,\underset{x\to 2}{\mathop{\lim }}\,\frac{{{x}^{2}}-4}{x-2}=4$ Now check whether $\,\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ or not. From the above, $\,\underset{x\to 2}{\mathop{\lim }}\,\frac{{{x}^{2}}-4}{x-2}=4\text{ and }f\left( 2 \right)=5$ Therefore, $\,\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)\ne f\left( 2 \right)$ 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}}-4}{x-2}\text{ if }x\ne 2 \\ & 5\text{ if }x=2 \end{align} \right. is not continuous at $2$.