limit and absolute absolute value problem $$\lim_{x \to -2} \frac{2-|x|}{2+x}$$
If I calculate the left and right-hand limit I get different results.
Left hand side:
$$\lim_{x \to -2^-}\frac{2+x}{2+x}=1$$
Right hand side: $$\lim_{x \to -2^+} \frac{2-x}{2+x}=\text{undefined}$$
My question is that my procedure is right or wrong?
 A: Since $x \to -2$, we can assume that $x < 0$ so that $|x| = -x$.
Then $$\frac{2-|x|}{2+x} = \frac{2+x}{2+x} = 1 \xrightarrow{x \to -2} 1$$
so the limit exists and it is equal to $1$.
A: When $x$ is near $-2$ approaching it from the left, $2+x$ is equivalent to $2-|x|$ (if $x<0$, then $x=-|x|$):
$$
\lim_{x\to-2^-}\frac{2-|x|}{2+x}=\lim_{x\to-2^-}\frac{2-|x|}{2-|x|}=\lim_{x\to-2^-}1=1.
$$
When $x$ is near $-2$ approaching it from the right, $2+x$ also seems to be equivalent to $2-|x|$:
$$
\lim_{x\to-2^+}\frac{2-|x|}{2+x}=\lim_{x\to-2^+}\frac{2-|x|}{2-|x|}=\lim_{x\to-2^+}1=1.
$$
Since both one-sided limits are equal to the same number, the limit exists and is equal to $1$:
$$\lim_{x\to-2}\frac{2-|x|}{2+x}=1.$$
Even though the function itself is undefined at $x=-2$ because the denominator at that point is zero ($f(-2)=\frac{2-|-2|}{2-2}=\frac{0}{0}$), the limit of this function at $x=-2$ does exist and is equal to $1$.
A: Multiply and divide by $2+|x|$ and cancel:
$$\lim_{x \to -2} \frac{2-|x|}{2+x}=\lim_{x \to -2} \frac{4-x^2}{(2+x)(2+|x|)}=\lim_{x \to -2} \frac{2-x}{2+|x|}=\frac{2-(-2)}{2+2}=1.$$
A: $$\lim_{x\rightarrow-2}\frac{2-|x|}{2+x}=\lim_{x\rightarrow-2}\frac{2+x}{2+x}=1.$$
A: What about L'Hospital's Rule?
Since 
$$\frac{2-|-2|}{2+-2}=\frac{0}{0}$$
then
$$\frac{d}{dx}(2-|x|)=-\frac{x}{|x|}, \frac{d}{dx}(2+x)=1$$
evaluate 
$\frac{-\frac{x}{|x|}}{1}$ 
at $x=-2$
yields
$$\frac{-\frac{-2}{|-2|}}{1} = \frac{-(-1)}{1}=1$$
