Answer
The one-sided limits are not equal and the limit does not exists.
Work Step by Step
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.
