Answer
$\dfrac{1}{6}$
Work Step by Step
Given: $\lim\limits_{x \to 3}\dfrac{x-3}{x^2-9}$
This shows an indeterminate form of type $\dfrac{0}{0}$, and so we will apply L'Hospital Rule.
$\lim\limits_{x\to3}\dfrac{x-3}{x^2-9}=\lim\limits_{x\to3}\frac{\dfrac{d(x-3)}{dx}}{\dfrac{d(x^2-9)}{dx}}$
or, $=\lim\limits_{x\to3}\dfrac{1}{2x}$
or,$=\dfrac{1}{2\times3}$
or, $=\dfrac{1}{6}$
