Answer
$\dfrac{-1}{4}$
Work Step by Step
Since, $\lim\limits_{x \to -2}f(x)=\lim\limits_{x \to -2}\dfrac{x+2}{x^2-4}$
Check for indeterminate form:
Thus, $f(-2)=\dfrac{-2+2}{4-4}=\dfrac{0}{0}$
Apply L-Hospital's rule:
$\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$
Thus, $\lim\limits_{x \to -2}\dfrac{1}{2x}=\dfrac{1}{2(-2)}=\dfrac{-1}{4}$
