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.

You can see that, for all $(x,y)$ in $\mathbb{R}^2-\{(0,0)\}$ :
$$\frac{x^3-2y^3}{x^2+2y^2}=\frac{x^3}{x^2+2y^2}-2\frac{y^3}{x^2+2y^2}$$
so, by triangular inequality :
$$\left|\frac{x^3-2y^3}{x^2+2y^2}\right|\le\frac{\vert x\vert^3}{x^2+y^2}+2\frac{\vert y\vert^3}{x^2+y^2}\le\vert x\vert+2\vert y\vert\le2(\vert x\vert+\vert y\vert)$$
It is now clear that the desired limit is $0$.
