## Precalculus (6th Edition) Blitzer

Discontinuous at $x=0$
Recall that if $f$ is a polynomial function, then we have $\lim_\limits{x\to a}f(x)=f(a)$. $\lim_\limits{x \to 0^{-}} f(x)= \lim_\limits{x \to 0^{+}}\dfrac{x^2+5x}{x^2-5x}=-1$ and $\lim_\limits{x \to 0^{+}} f(x)= \lim_\limits{x \to 0^{+}}\dfrac{x^2+5x}{x^2-5x}=-1$ So, $\lim_\limits{x \to 0^{-}} f(x)= \lim_\limits{x \to 0^{+}}f(x)$ exists. Now, $\lim_\limits{x \to 0^{-}} f(x)=-1$ and $f(0)=-2$ so, $\lim_\limits{x \to 0} f(x) \neq f(0)$ Therefore, the function is Discontinuous at $x=0$