Answer
1
Work Step by Step
Since $|x|=-x$ for $x<0$, we have
$|x|=\left\{\begin{array}{lll}
x & if & x \geq 0\\
-x & if & x < 0
\end{array}\right.$
Approaching $x=-2$ (from either side, $x < 0$)
$\displaystyle \lim_{x\rightarrow-2}\frac{2-|x|}{2+x}=\lim_{x\rightarrow-2}\frac{2-(-x)}{2+x}$
$=\displaystyle \lim_{x\rightarrow-2}\frac{(2+x)}{(2+x)}$
$=\displaystyle \lim_{x\rightarrow-2}1$
$=1$