Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x\to-2}\frac{2-|x|}{2+x}=1$
$A=\lim\limits_{x\to-2}\frac{2-|x|}{2+x}$ We see that $$2-|x|=\left\{ \begin{array} {c l} 2-x && x\geq0\\ 2-(-x)=2+x && x<0 \end{array} \right.$$ In this case, we try to find the limit of the function as $x$ approaches $-2$. Therefore, we only care about the neighbourhood value of $-2$, which is the values very near to $-2$. In other words, the values of $x\geq0$ are not considered because they are too far from $-2$. So, $2-|x|=2+x$ as $x\lt0$ Which means, $A=\lim\limits_{x\to-2}\frac{2+x}{2+x}$ $A=\lim\limits_{x\to-2}1$ $A=1$