## Precalculus (6th Edition) Blitzer

Recall that if $f$ is a polynomial function, then we have $\lim_\limits{x\to a}f(x)=f(a)$. $\lim_\limits{x \to 5^{-}} f(x)= \lim_\limits{x \to 5^{-}} 4x=20$ and $\lim_\limits{x \to 5^{+}} f(x)= \lim_\limits{x \to 5^{+}}(x^2-5)=20$ So, $\lim_\limits{x \to 5^{-}} f(x)= \lim_\limits{x \to 5^{+}}f(x)$ exists. Now, $\lim_\limits{x \to 5} f(x)=20$ and $f(5)=20$ so, $\lim_\limits{x \to 5} f(x) = f(5)$ Therefore, the function is continuous everywhere.