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?

These things are always nicer in polar form... 
$$\frac{x^4+y^4}{x^2+y^2} = \frac{r^4 (\sin^4 \theta + \cos^4 \theta)} {r^2} =  r^2  (\sin^4 \theta + \cos^4 \theta) \leq 2r^2$$
Let's assume $\|(x,y)\| < \delta$, that is $(x,y)$ is inside the open disk around $0$ with a radius of $\delta$.
Then using the above inequality we can guarantee that 
$$\left\| \frac{x^4+y^4}{x^2+y^2} \right\| \leq 2\delta^2,$$
which in turn means that the image of $(x,y)$ is inside the open disk around $0$ with a radius of $2\delta^2$.
Therefore if you get an $0 < \epsilon$, and you pick $\delta$, such that $2\delta^2 < \epsilon$, then
$$\|(x,y)\| < \delta \Rightarrow \left\| \frac{x^4+y^4}{x^2+y^2} \right\| \leq 2\delta^2 < \epsilon.$$
