Answer
$-4$
Work Step by Step
Consider $f(x)=\lim\limits_{x \to 0}\dfrac{4-4e^x}{xe^x}=\lim\limits_{x \to 0} \dfrac{a(x)}{b(x)}$ and $a(0)=0, b(0)=0$
Thus, $f(0)=\dfrac{0}{0}$
This shows an Inderminate form of the limit, so apply L-Hospital's rule:
$\lim\limits_{x \to l}\dfrac{a(x)}{b(x)}=\lim\limits_{x \to l}\dfrac{a'(x)}{b'(x)}$
Thus,
$\lim\limits_{x \to 0}\dfrac{-4e^x}{e^x+xe^x}=\lim\limits_{x \to 0}\dfrac{-4e^0}{e^0+0}=-4$
