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:
$$| \frac{x^2y}{x^2+y^3} | = |y-\frac{y^4}{x^2+y^3}| \le$$ 
 A: Note that for $x \neq 0$ and $y \neq 0$ we have
$$ \left | \frac{x^2y}{x^2 + y^3} \right| = \left| \frac{y}{1 + \frac{y^3}{x^2}} \right| = \left| \frac{1}{\frac{1}{y} + \frac{y^2}{x^2}} \right| \leq \left| \frac{1}{\frac{1}{y}} \right| = \left| y \right|. $$
If $x = 0$ or $y = 0$ (but not $x = y = 0$), then we also have
$$ \left | \frac{x^2y}{x^2 + y^3} \right| = 0 \leq |y|. $$
Thus, by definition or by squeeze theorem, we have
$$ \lim_{(x,y) \rightarrow (0,0)} \frac{x^2y}{x^2 + y^3} = 0. $$
ERRATA: Let $f(x,y) = \frac{x^2y}{x^2 + y^3}$. Note that in the "solution" above, in the first line, we have used $y > 0$ for the inequality to be true. Indeed, if we restrict our function to the upper half plane, then the limit of the function is zero. However, if we consider $f$ as a function that is defined on its natural domain of definition $\mathbb{R}^2 \setminus \{ (x,y) \, | \, x^2 + y^3 \neq 0 \}$, then the function doesn't have a limit as $(x,y) \to (0,0)$. You can see this if you plot the graph of the function. Analytically,
$$ \lim_{n \to \infty} f \left( \frac{1}{n^3}, -\frac{1}{n^2} + \frac{1}{n^4} \right) = \frac{\frac{1}{n^6} \left( -\frac{1}{n^2} + \frac{1}{n^4} \right)}{\frac{1}{n^6} + \left(-\frac{1}{n^2} + \frac{1}{n^4} \right)^3} = \frac{\frac{1}{n^{10}} - \frac{1}{n^6}}{\frac{3}{n^8} - \frac{3}{n^{10}} + \frac{1}{n^{12}}} \\
 = \lim_{n \to \infty} \frac{\frac{1}{n^4} - 1}{\frac{3}{n^2} - \frac{3}{n^4} + \frac{1}{n^6}}  = -\infty.$$
A: Let $a>b>0$, $x=a\cdot t$ and $b\cdot t$. Then we have 
$$
\lim_{t^+\to 0}\frac{a^2t^2\cdot bt}{a^2t^2 + b^3t^3}
=
\lim_{t^+\to 0}\frac{a^2bt^3}{t^3(a^2\frac{1}{t} + b^3)}
=
\lim_{t^+\to 0}\frac{a^2b}{(a^2\frac{1}{t} + b^3)}=+\infty
$$
and
$$
\lim_{t^-\to 0}\frac{a^2t^2\cdot bt}{a^2t^2 + b^3t^3}
=
\lim_{t^-\to 0}\frac{a^2bt^3}{t^3(a^2\frac{1}{t} + b^3)}
=
\lim_{t^-\to 0}\frac{a^2b}{(a^2\frac{1}{t} + b^3)}=-\infty
$$
A: 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.
