## Calculus (3rd Edition)

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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 $$\lim\limits_{x \to 3} (x^2-9)\frac{x-3}{|x-3|}=0.$$