Proving that a limit does not exist with absolute values I'm to prove that the following limit does not exist
$\lim_{x\to -2} \frac{\vert 2x +4\vert -\vert x^3 +8 \vert}{x+2}$
From here, I have taken the method of finding $\lim_{x\to -2^+}$ and $\lim_{x\to -2^-}$ to show that they're not equal
However my problem is that both are simplified to become $\frac{x^3 - 2x +4}{x+2}$
Is there something I'm doing wrong or did i make a mistake when opening the absolutes?

Because for $x\rightarrow-2+$ we have
$$\frac{|2x+4|-|x^3+8|}{x+2}\rightarrow2-4-4-4=-10$$ and
for $x\rightarrow-2-$ we have
$$\frac{|2x+4|-|x^3+8|}{x+2}\rightarrow-2+4+4+4=10.$$
