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?

Take any $\epsilon > 0$. Suppose that $\|(x,y) - (0,0)\|<\delta(\epsilon) = \sqrt{\epsilon/2}$, which is to say that $x^2 + y^2 <\epsilon/2$.  We have
$$
\left|\frac{x^4 + y^4}{x^2 + y^2} - 0 \right| = 
\left|\frac{x^4}{x^2 + y^2} + \frac{y^4}{x^2 + y^2}\right| \leq
\left|\frac{x^4}{x^2 + y^2}\right| + \left|\frac{y^4}{x^2 + y^2}\right|\leq\\
\frac{(x^2 + y^2)^2}{x^2 + y^2} + \frac{(x^2 + y^2)^2}{x^2 + y^2} \leq\\
2(x^2 + y^2) < \epsilon
$$
