$\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? 

$$
\lim_{x\rightarrow -8} \frac{8-|x|}{8+x} = \lim_{x<0,\ x\rightarrow
-8} \frac{8+x}{8+x}=1 $$
