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?

Your simplification is only correct for one side. If you include more steps, we can pinpoint what went wrong
For $x > -2$
$$ \frac{|2x+4|-|x^3+8|}{x+2} = \frac{(2x+4)-(x^3+8)}{x+2} = \frac{(x+2)(2 - (x^2+2x+4))}{x+2} \\ = -(x^2+2x+2) $$
For $x < -2$
$$ \frac{|2x+4|-|x^3+8|}{x+2} = \frac{-(2x+4)+(x^3+8)}{x+2} = \frac{(x+2)(-2 + (x^2+2x+4))}{x+2} \\ = x^2+2x+2 $$
You can work out the two limits from here.
