Help find this limit I have the limit below 

$$\lim_{(x,y)\rightarrow(0,0)}\frac{x^3-2y^3}{x^2+2y^2}$$

I know that the limit must be zero if it exists since coming along the line $y=mx$ for an constant $m$  gives $0$, but I don't know how to prove it. I want to use squeeze theorem but I don't know what function to use for it.

Approach with ellipses $x=r \cos (\theta)$ and $y=\frac{1}{\sqrt{2}}r \sin (\theta)$ with $r \to 0^+$ regardless of $\theta$.
This gives,
$$\lim_{r \to 0^+} \frac{r^3 \cos^3(\theta)-\frac{1}{\sqrt{2}}r^3 \sin^3 (\theta)}{r^2}$$
$$=\lim_{r \to 0^+} r(\cos^3 (\theta)-\frac{1}{\sqrt{2}}\sin^3 (\theta))$$
$$=0$$
Or note,
$$|\frac{x^3}{x^2+2y^2}| \leq |\frac{x^3}{x^2}|$$
$$=|x| \to 0$$
Similarly we have,
$$|\frac{-2y^3}{x^2+2y^2 }| \leq |\frac{2y^3}{2y^2}|$$
$$=|y| \to 0$$
Now use $|a+b| \leq |a|+|b|$ to conclude with squeeze.
