Answer
$$0$$
Work Step by Step
Since $-1\leq \frac{x-3}{|x-3|}\leq 1$, then we have
$$-(x^2-9)\leq (x^2-9)\frac{x-3}{|x-3|}\leq (x^2-9).$$
Moreover, $\lim\limits_{x \to 3}(x^2-9)=\lim\limits_{x \to 3}-(x^2-9)=0$. Then by the Squeeze Theorem, we have
$$\lim\limits_{x \to 3} (x^2-9)\frac{x-3}{|x-3|}=0.$$
