$\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? 

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. 
