## Precalculus (6th Edition) Blitzer

No, the function $f\left( x \right)=\frac{{{x}^{2}}+8x}{{{x}^{2}}-8x}$ is not continuous at $0$.
Consider the function $f\left( x \right)=\frac{{{x}^{2}}+8x}{{{x}^{2}}-8x}$, First check whether the function is defined at the point $a$ or not. Find the value of $f\left( x \right)$ at $a=0$, $f\left( 0 \right)=\frac{{{\left( 0 \right)}^{2}}+8\left( 0 \right)}{{{\left( 0 \right)}^{2}}-8\left( 0 \right)}$ The function is not defined at the point $0$. Thus, the function does not satisfy the first property of being continuous. Hence, the function $f\left( x \right)=\frac{{{x}^{2}}+8x}{{{x}^{2}}-8x}$ is not continuous at $0$.