Prove a limit with $\varepsilon$-$\delta$ definition: the limit we want to prove is:
$$\lim_{(x,y)\to(0,0)}\frac{x^4}{x^2+y^2}=0$$
It is actually very simple: the denominator is the square of the norm of $(x,y)$, thus it follows that, for every $\varepsilon>0$ we must find $\delta>0$ for which it holds:
$$\frac{x^4}{\delta^{2}}<\varepsilon$$
but I do not see how to get rid of that $x^4$ in order to define a proper $\delta$. I imagine one has to use some particular inequality based on squares I must have forgot.

$$0\le \frac{x^4}{x^2+y^2}=x^2\frac{x^2}{x^2+y^2}\le x^2\le x^2+y^2=\|(x,y)\|^2$$
So just take $\delta=\sqrt{\varepsilon}$
