Answer
$$\lim_{x\to3}\frac{x-3}{x^2-9}=\frac{1}{6}$$
Work Step by Step
$$A=\lim_{x\to3}\frac{x-3}{x^2-9}$$
In this exercise, the L'Hospital's Rule is actually not necessary. However, for the sake of everybody, I will carry out both methods, elementary one and L'Hospital's Rule one.
1) Method 1: Elementary method
$$A=\lim_{x\to3}\frac{x-3}{x^2-9}$$
$$A=\lim_{x\to3}\frac{x-3}{(x-3)(x+3)}$$
$$A=\lim_{x\to3}\frac{1}{x+3}$$
$$A=\frac{1}{3+3}$$
$$A=\frac{1}{6}$$
2) Method 2: L'Hospital's Rule
Since $\lim_{x\to3}(x-3)=3-3=0$ and $\lim_{x\to3}(x^2-9)=3^2-9=0$,
this limit is an indeterminate form of type $\frac{0}{0}$, and we can apply L'Hospital Rule
$$A=\lim_{x\to3}\frac{x-3}{x^2-9}$$
$$A=\lim_{x\to3}\frac{\frac{d(x-3)}{dx}}{\frac{d(x^2-9)}{dx}}$$
$$A=\lim_{x\to3}\frac{1}{2x}$$
$$A=\frac{1}{2\times3}$$
$$A=\frac{1}{6}$$
