Proving $\lim_{(x,y)\to (0,0)} (x^4+y^4)/(x^2+y^2)=0$ by definition I need to prove by definition (and nothing else) that
$$\lim_{(x,y) \to (0,0)}\frac{x^4+y^4}{x^2+y^2} = 0.$$
I've been stuck on this for almost an hour with no luck, and ran out of ideas.
Can anyone help or give a hint?

Another approach which is useful whenever you see a denominator of $x^2+y^2$ is to use polars: $x=r\cos t$, $y=r\sin t$. Then
$$\frac{x^4+y^4}{x^2+y^2}=\frac{r^4\cos^4t+r^4\sin^4t}
{r^2\cos^2t+r^2\sin^2t}=r^2(\sin^4t+\cos^4t)$$
which is clearly $\le 2r^2$. So as $r\to0$ the function tends to zero.
(Although as Luiz points out an even
stronger inequality is obvious.)
