[tex]\displaystyle\lim_{x\to-8}\frac{5x+40}{|x+8|}=5\lim_{x\to-8}\dfrac{x+8}{|x+8|}[/tex]
Recall that
[tex]|x+8|=\begin{cases}x+8&\text{for }x+8\ge0\\-(x+8)&\text{for }x+8<0\end{cases}[/tex]
For the limit to exist, the limits from either side must also exist and match up. When [tex]x\to-8^-[/tex], we're considering the domain [tex]x<-8[/tex], which corresponds to the second case in the piecewise definition above, i.e. [tex]|x+8|=-(x+8)[/tex]. So we have
[tex]\displaystyle\lim_{x\to-8^-}\frac{x+8}{|x+8|}=\lim_{x\to-8}\frac{x+8}{-(x+8)}=-1[/tex]
On the other hand, as [tex]x\to-8^+[/tex], we're assuming [tex]x>-8[/tex], so that [tex]|x+8|=x+8[/tex], in which case
[tex]\displaystyle\lim_{x\to-8^+}\frac{x+8}{|x+8|}=\lim_{x\to-8}\frac{x+8}{x+8}=1[/tex]
The limits do not match because they differ by their sign, so the original limit does not exist.
