## Precalculus (6th Edition) Blitzer

Yes, the function $f\left( x \right)=\frac{x-5}{x+5}$ is continuous at $5$.
Consider the function $f\left( x \right)=\frac{x-5}{x+5}$, First check whether the function is defined at the point $a$ or not. Find the value of $f\left( x \right)$ at $a=5$, \begin{align} & f\left( 5 \right)=\frac{5-5}{5+5} \\ & =0 \end{align} The function is defined at the point $a=5$. Now find the value of $\,\underset{x\to 5}{\mathop{\lim }}\,\frac{x-5}{x+5}$, \begin{align} & \underset{x\to 5}{\mathop{\lim }}\,f\left( x \right)\,=\underset{x\to 5}{\mathop{\lim }}\,\frac{x-5}{x+5} \\ & =\frac{\,\underset{x\to 5}{\mathop{\lim }}\,\left( x-5 \right)}{\,\underset{x\to 5}{\mathop{\lim }}\,\left( x+5 \right)}\, \\ & =\frac{5-5}{5+5} \\ & =0 \end{align} Thus, $\,\underset{x\to 5}{\mathop{\lim }}\,\frac{x-5}{x+5}=0$ Now check whether $\,\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ or not. From the above, $\,\underset{x\to 5}{\mathop{\lim }}\,\frac{x-5}{x+5}=0\text{ and }f\left( 5 \right)=0$ Therefore, $\,\underset{x\to 5}{\mathop{\lim }}\,f\left( x \right)=f\left( 5 \right)$ Thus, the function satisfies all the properties of being continuous. Hence, the function $f\left( x \right)=\frac{x-5}{x+5}$ is continuous at $5$.