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?
 A: 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.)
A: 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
$$
\left|\frac{x^4 + y^4}{x^2 + y^2} - 0 \right| = 
\left|\frac{x^4}{x^2 + y^2} + \frac{y^4}{x^2 + y^2}\right| \leq
\left|\frac{x^4}{x^2 + y^2}\right| + \left|\frac{y^4}{x^2 + y^2}\right|\leq\\
\frac{(x^2 + y^2)^2}{x^2 + y^2} + \frac{(x^2 + y^2)^2}{x^2 + y^2} \leq\\
2(x^2 + y^2) < \epsilon
$$
A: Note that $\frac{x^4+y^4}{x^2+y^2}\leq x^2+y^2$. The rest should be obvious.
A: 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
$$\|(x,y)\| < \delta \Rightarrow \left\| \frac{x^4+y^4}{x^2+y^2} \right\| \leq 2\delta^2 < \epsilon.$$
