Computing $\lim_{x\to-5}\frac{x^2+2x-15}{|x+5|}$ The problem is: $$\lim_{x\to-5}\frac{x^2+2x-15}{|x+5|}$$
I factored the numerator to get: $\frac{(x-3)(x+5)}{|x+5|}$
How do i solve the rest?

Another place where "factor out the big part" helps.  Here, $|x|$ is growing without bound so $x^2$ is the big part in the numerator and $|x|$ is the big part in the denominator.
\begin{align*}
L &= \lim_{x \rightarrow -\infty} \frac{x^2 + 2x - 15}{|x+5|}   \\
&= \lim_{x \rightarrow -\infty} \frac{x^2(1 + 2/x - 15/x^2)}{|x(1+5/x)|}   \\
    &= \lim_{x \rightarrow -\infty} \frac{x^2}{|x|} \frac{1 + 2/x - 15/x^2}{|1+5/x|}
\end{align*}
We should be able to see that the second fraction is going to $\frac{1+0-0}{|1+0|}$.  (Making this easy to see is why you factor out the "big part".)  The first fraction is always positive (positive over positive) once $x < 0$, so
$$ L = \lim_{x \rightarrow -\infty} (|x| \cdot 1) = \infty  \text{.}  $$
