## Precalculus (6th Edition) Blitzer

Discontinuous at $x=-3$
Recall that if $f$ is a polynomial function, then we have $\lim_\limits{x\to a}f(x)=f(a)$. $f(x)=\dfrac{x^2-9}{x+3}$ Set $x=-3$ $f(-3)=\dfrac{(-3)^2-9}{-3+3}=0$ So, $f(-3)$ is not defined. Therefore, the function is discontinuous at $x=-3$