$\displaystyle\lim_{x\to -8} \frac {8 − |x|} {8 + x} \ $ How do I solve? \begin{align*}
\lim_{x\to -8^+} \frac {8 − |x|} {8 + x} &=
\lim_{x\to -8^+} \frac {8 − x} {8 + x}\\
&= \frac {8 − 8} {8 + -8} \\ &= 0
\end{align*}
\begin{align*}
\lim_{x\to -8^-} \frac {8 − |x|} {8 + x} &=
\lim_{x\to -8^-} \frac {8 − (-x)} {8 + x} \\
&= \frac {8 + − 8} {8 + -8} \\ &= 0 
\end{align*}
However, the answer key says that I am wrong. Where have I failed? 
 A: $$
\lim_{x\rightarrow -8} \frac{8-|x|}{8+x} = \lim_{x<0,\ x\rightarrow
-8} \frac{8+x}{8+x}=1 $$
A: Note that $\frac{8-|x|}{8+x}=\frac{8-x}{8+|x|}$, so $$\lim_{x\rightarrow-8}\frac{8-|x|}{8+x}=\lim_{x\rightarrow-8}\frac{8-x}{8+|x|}=\frac{\lim_{x\rightarrow-8}(8-x)}{\lim_{x\rightarrow-8}(8+|x|)}=\frac{16}{16}=1$$.
Note that $\lim_{x\rightarrow-8}(8+|x|)=16\neq0$, so the above works.
A: Specifically with respect to your answer, the biggest mistake is obviously in
$$\frac {8 + 8} {8 + -8} = 0 \quad\text{and}\quad \frac {8 + -8} {8 + -8} = 0$$
Here you have $0$ in the denominator, which makes both fractions undefined. Moreover, whether you approach $-8$ from the left or from the right, you are still dealing with a negative number in a small enough neighborhood around $-8$, so you may assume that $x<0$. The correct result, as have been suggested by other answerers, is
$$\lim_{x\to-8} \frac{8-\vert x\vert}{8+x}=\lim_{x\to -8}\frac{8+x}{8+x}=1$$
A: Let $\alpha$ stand for a number infinitesimally close to 0, but not equal to 0.
Now, we have \begin{align*}
\lim_{x\to -8^+} \frac {8 − |x|} {8 + x} &
\end{align*} 
becomes
\begin{align*}
\frac {8 − |-8+\alpha|} {8 + -8+\alpha} &
\end{align*}.
Since |-8+$\alpha$|=8, we thus end up with $\alpha$/$\alpha$=1, for the first limit.  Similarly reasoning yields 1 for the second limit. 
