How to show that $\lim_{(x,y) \to (0,0)}\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}=0$ How to show that $$\lim_{(x,y) \to (0,0)}\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}=0$$
I know that
$$\left|\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}\right|\le\frac{x^{2}y^{2}}{\left|x\right|^{3}}=\frac{y^{2}}{\left|x\right|}$$
But now we have a problem with $\left|x\right|$ and I cannot find a lower bound other than $0$.
Also$$\left|\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}\right|\le\frac{x^{2}y^{2}}{\left|x^{3}+y^{3}\right|}=\frac{\left|x+y\right|}{\left|x^{2}-xy+y^{2}\right|}\le\frac{\left|x\right|+\left|y\right|}{\left|x^{2}-xy+y^{2}\right|}$$
But I'm not sure if that helps.
If we show that $\lim_{(x,y) \to (0,0)}\frac{y^{2}}{\left|x\right|}=0$ then we are done.
 A: You can directly apply the general statement:
If $a, b \ge 0$, $c, d > 0$ and $\frac{a}{c} + \frac{b}{d} > 1$ then
$$\lim_{(x,y)\rightarrow (0,0)}\frac{\left|x\right|^a\left|y\right|^b}{\left|x\right|^c + \left|y\right|^d} = 0$$
You can also work it out in an ad hoc way.
Let $u=|x|^3$ and $v=|y|^3$
and note that
$$
u^{2/3}v^{2/3}=u^{1/3}(u^{1/3}v^{2/3})\le u^{1/3}(\frac13u+\frac23 v)\tag{1}
\le u^{1/3}(u+v)
$$
and thus
$$
|f(x,y)|=\frac{u^{2/3}v^{2/3}}{u+v}\le u^{1/3}=|x|
$$
In the first inequality of (1), Young's inequality is used.
A: Easier: if $(x,y) = (r\cos\theta,r\sin\theta)$,
$$
\left|\frac{x^2 y^2}{|x|^3+|y|^3}\right|
= \frac{r^4\cos^2\theta\sin^2\theta}{r^3(|\cos\theta|^3 + |\sin\theta|^3)}\le\frac{r}{m} = \frac{\sqrt{x^2 + y^2}}{m},
$$
with
$$m = \min_{\theta\in[0,2\pi]}(|\cos\theta|^3 + |\sin\theta|^3) > 0.$$
(Why $> 0$)?
A: AM-GM:
$|x|^3 +|y|^3 \ge 2|x|^{3/2}|y|^{3/2};$
$\dfrac {x^2y^2}{|x|^3+|y|^3} \le \dfrac {x^2y^2}{2|x|^{3/2}|y|^{3/2}}=$
$(1/2)|x|^{1/2}|y|^{1/2} \rightarrow 0.$
