Answer
$-4$
Work Step by Step
Consider $f(x)=x^2; x=-2$
and $f(x+h)=(x+h)^2=x^2+2xh+h^2$
Therefore,
$\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}=\lim\limits_{h\to0}\dfrac{(x^2+2xh+h^2)-(x^2)}{h}=\lim\limits_{h\to0}\dfrac{(2x)h+h^2}{h}=\lim\limits_{h\to0}(2x+h)$
Plug $x=-2$, we get
$\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}=\lim\limits_{h\to 0}(-4+h)=-4$$
