Computing the limit of a 2-variable function Show that $\lim \limits _{(x,y)\rightarrow (0,0)} \frac{x^3y}{x^2+y^4}=0$ just using $\epsilon-\delta$ creterion.
In fact, choose $\epsilon >0$ arbitary, then we have to find $\delta >0$ such that if $\sqrt{x^2+y^2}<\delta$ then $\left\vert \frac{x^3y}{x^2+y^4}\right\vert <\epsilon$.  

Hint: $$\forall (x,y)\in \mathbb R^2\left((x,y)\neq (0,0)\implies \left|\dfrac{x^3y}{x^2+y^4}\right|\leq |xy|\leq \dfrac{x^2+y^2}2=\dfrac {\left\Vert (x,y)-(0,0)\right\Vert^2}2\right).$$
Answer:
The goal is to prove that $$\mathop{\forall}_{\varepsilon >0}\mathop{\exists}_{\delta >0}\mathop{\forall}_{(x,y)\in \mathbb R^2\setminus \{(0,0)\}}\left(\left\Vert(x,y)-(0,0) \right\Vert<\delta \implies  \left|\dfrac{x^3y}{x^2+y^4}\right|<\varepsilon\right).$$
Take an arbitrary $\varepsilon>0$, set $\delta=\sqrt{2\varepsilon}$ and let $(x,y)\in \mathbb R^2$ with $(x,y)\neq (0,0)$ be given.
If $\left\Vert(x,y)-(0,0) \right\Vert<\delta$, from the hint it follows that $\left|\dfrac{x^3y}{x^2+y^4}\right|<\dfrac{\delta^2}2=\varepsilon$, as desired.
