Proving that a limit does not exist with absolute values

Question

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?

Answer 1

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.

Answer 2

Hint:

$$\frac{|2x+4|-|x^3+8|}{x+2}=\frac{2|x+2|-|x+2||x^2-2x+4|}{x+2}=\frac{|x+2|}{x+2}\left(-x^2+2x-2\right)$$

Answer 3

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 $$\frac{|2x+4|-|x^3+8|}{x+2}\rightarrow-2+4+4+4=10.$$

Answer 4

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.

Answer 5

Rewrite the expression as

$$\frac{2|x+2|}{x+2}-\frac{|x+2||x^2-2x+4|}{x+2}=\frac{|x+2|}{x+2}(2-|x^2-2x+4|).$$

Clearly, the first factor tends to $\pm1$ depending on the side, while the second has the limit $-10$.