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?

Let $x=y-2$ with $y\to0$ then
$$\frac{\vert 2x +4\vert -\vert x^3 +8 \vert}{x+2}=\frac{\vert 2y\vert -\vert y^3+3y^2+3y\vert}{y}=\frac{\vert 2y\vert -\vert y^3-6y^2+12y\vert}{y}=\frac{|y|}{y}(2-|y^2-6y+12|)\to\pm1\cdot(2-12)=\pm10 $$
thus the limit does not exist.
