## Calculus 8th Edition

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, $x < 0$) $\displaystyle \lim_{x\rightarrow-2}\frac{2-|x|}{2+x}=\lim_{x\rightarrow-2}\frac{2-(-x)}{2+x}$ $=\displaystyle \lim_{x\rightarrow-2}\frac{(2+x)}{(2+x)}$ $=\displaystyle \lim_{x\rightarrow-2}1$ $=1$