## Calculus (3rd Edition)

We have $$\lim _{x \rightarrow 3}\frac{x^2-x}{x^2-9}=\lim _{x \rightarrow 3}\frac{x(x-1)}{(x+3)(x-3)}=\frac{6}{0}$$ hence the limit does not exist. The one-sided limits are, $$\lim _{x \rightarrow 3^+}\frac{x^2-x}{x^2-9}=\lim _{x \rightarrow 3^+}\frac{x(x-1)}{(x+3)(x-3)}=\frac{6}{0}=\infty$$ $$\lim _{x \rightarrow 3^-}\frac{x^2-x}{x^2-9}=\lim _{x \rightarrow 3^-}\frac{x(x-1)}{(x+3)(x-3)}=-\frac{6}{0}=-\infty$$ hence the one-sided limits are not equal and the limit does not exist.