## Precalculus (6th Edition) Blitzer

Yes, the function $f\left( x \right)=\frac{{{x}^{2}}+4}{x-2}$ is continuous at $3$.
Consider the function $f\left( x \right)=\frac{{{x}^{2}}+4}{x-2}$, First check whether the function is defined at the point $a$ or not. Find the value of $f\left( x \right)$ at $a=3$, \begin{align} & f\left( 3 \right)=\frac{{{\left( 3 \right)}^{2}}+4}{\left( 3 \right)-2} \\ & =\frac{9+4}{3-2} \\ & =13 \end{align} The function is defined at the point $a=3$. Now find the value of $\,\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}+4}{x-2}$, \begin{align} & \,\underset{x\to 3}{\mathop{\lim }}\,f\left( x \right)=\,\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}+4}{x-2} \\ & =\frac{\,\underset{x\to 3}{\mathop{\lim }}\,\left( {{x}^{2}}+4 \right)}{\,\underset{x\to 3}{\mathop{\lim }}\,\left( x-2 \right)}\, \\ & =\frac{{{\left( 3 \right)}^{2}}+4}{3-2} \\ & =13 \end{align} Thus, $\,\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}+4}{x-2}=13$ Now check whether $\,\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ or not. From the above, $\,\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}+4}{x-2}=13\text{ and }f\left( 3 \right)=13$ Therefore, $\,\underset{x\to 3}{\mathop{\lim }}\,f\left( x \right)=f\left( 3 \right)$ Thus, the function satisfies all the properties of being continuous. Hence, the function $f\left( x \right)=\frac{{{x}^{2}}+4}{x-2}$ is continuous at $3$.