Answer
$\lim\limits_{x \to 3}\frac{x-3}{x^2-9}= \frac{1}{6}$
Work Step by Step
From the graph we see that $\lim\limits_{x \to 3}\frac{x-3}{x^2-9}$ approaches the value $0.1666666$.... which is approximately equal to $\frac{1}{6}$
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