Answer
$\dfrac{-1}{4}$
Work Step by Step
Consider: $\lim\limits_{x \to -2}f(x)=\lim\limits_{x \to -2}\dfrac{x+2}{x^2-4}$
Need to check that the limit has an indeterminate form.
Thus, $f(-2)=\dfrac{-2+2}{4-4}=\dfrac{0}{0}$
Now, apply L-Hospital's rule such as: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$
This implies:
$\lim\limits_{x \to -2}\dfrac{1}{2x}=\dfrac{1}{2(-2)}=\dfrac{-1}{4}$
