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.
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 $0$.