Find multivariable limit $\frac{x^2y}{x^2+y^3}$ Find multivariable limit of: $$\lim_{ \left( x,y\right) \rightarrow \left(0,0 \right)}\frac{x^2y}{x^2+y^3}$$
How to find that limit? I was trying to do the following, but i am not able to find a proper inequality:
$$| \frac{x^2y}{x^2+y^3} | = |y-\frac{y^4}{x^2+y^3}| \le$$ 

Late answer since it was marked as duplicate somewhere else but the way I suggest was not among the answers.
Looking at the denominator of $\frac{x^2y}{x^2+y^3}$ it makes sense to consider a path of approaching $(0,0)$ close to the curve $x^2+y^3=0$.
So, consider for $x> 0$ and $\alpha > 0$ the path
$$(x,-\sqrt[3]{x^2 + x^{\alpha}})\Rightarrow \frac{x^2y}{x^2+y^3}=\frac{x^2\sqrt[3]{x^2 + x^{\alpha}}}{x^{\alpha}}\geq \frac{x^{2+\frac{\alpha}{3}}}{x^{\alpha}} = \frac{1}{x^{\frac{2}{3}\alpha-2}}$$
For $\frac{2}{3}\alpha-2 >0 \Leftrightarrow \alpha > 3$ we get
$$\frac{1}{x^{\frac{2}{3}\alpha-2}}\stackrel{x \to 0^+}{\longrightarrow}+\infty$$
So, approaching $(0,0)$ along such a path shows that the limit does not exist.
