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?

After the correction in the limit $x\rightarrow -5$.
Recall that $|x+5|=\begin{cases}x+5 \space\space \text{if $x\geq-5$} \\-(x+5)\space\space \text{if $x<-5$}\end{cases}.$ Then

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*$\lim_{x\rightarrow (-5)^{-}}\frac{(x-3)(x+5)}{|x+5|}=\lim_{x\rightarrow (-5)^{-}}\frac{(x-3)(x+5)}{-(x+5)}=\lim_{x\rightarrow (-5)^{-}}-(x-3)=-(-5-3)=8$ since we approach from $x<-5$

*$\lim_{x\rightarrow (-5)^{+}}\frac{(x-3)(x+5)}{|x+5|}=\lim_{x\rightarrow (-5)^{+}}\frac{(x-3)(x+5)}{(x+5)}=\lim_{x\rightarrow (-5)^{+}}(x-3)=-5-3=-8$ since we approach from $x>-5.$
Thus the limit doesn't exist.
