Limit with polar coordinates $ \lim_{(x,y) \to (0,0)} \frac{x^3y}{x^4+y^2} = 0$ I need to prove that 
$$ \lim_{(x,y) \to (0,0)} \frac{x^3y}{x^4+y^2} = 0$$
I'm using polar coordinates
$$\lim_{\rho \to 0} \frac{\rho^4\cos^3(\theta)\sin(\theta)}{\rho^2(\rho^2\cos^4(\theta)+\sin^2(\theta))} = \lim_{\rho \to 0} \frac{\rho^2\cos^3(\theta)\sin(\theta)}{\rho^2\cos^4(\theta)+\sin^2(\theta)} $$
From that point I can't find a function $g(\rho)\to 0$ for $\rho\to0$. 
$$|\frac{\rho^2\cos^3(\theta)\sin(\theta)}{\rho^2\cos^4(\theta)+\sin^2(\theta)}|\leq\frac{\rho^2}{\rho^2\cos^4(\theta)+\sin^2(\theta)}$$
This is where I get stuck. Thank you in advance for your help! 
 A: I would use the AM-GM inequality:
$$x^4+y^2\geq 2x^2|y|$$ so
$$\frac{|x|^3|y|}{x^4+y^2}\le \frac{|x^3||y|}{2x^2|y|}=\frac{|x|}{2}$$ and this tends to zero.
A: The proof with polar is messy, but I will finish it, and then provide an alternative approach.

Polar Approach:
The key is to consider the cases when $\sin^2(\theta)<\rho$ and $\sin^2(\theta)\ge\rho$.
When $\sin^2(\theta)<\rho$ we have
$$\frac{\rho^2|\cos^3(\theta)\sin(\theta)|}{\rho^2\cos^4(\theta)+\sin^2(\theta)}<\frac{\rho^3|\cos^3(\theta)|}{\rho^2\cos^4(\theta)}=\frac\rho{\sqrt{1-\sin^2(\theta)}}<\frac\rho{\sqrt{1-\rho}}\to0$$
When $\sin^2(\theta)\ge\rho$ we have
$$\frac{\rho^2|\cos^3(\theta)\sin(\theta)|}{\rho^2\cos^4(\theta)+\sin^2(\theta)}<\frac{\rho^2}{\sin^2(\theta)}\le\frac{\rho^2}\rho=\rho\to0$$
Hence the limit is $0$, independent of $\theta$.

Alternative Approach:
Polar relies on $\rho^2=x^2+y^2$, which, in this case, results in a fairly messy expression. Alternatively, we can let $\rho^2=x^4+y^2$ to match our denominator. We thus have the inequalities $\rho^2\ge x^4$ and $\rho^2\ge y^2$, or $|x|\le\sqrt\rho$ and $|y|\le\rho$ (since $\rho>0$). Then we have
$$\frac{|x^3y|}{x^4+y^2}\le\frac{(\sqrt\rho)^3\rho}{\rho^2}=\sqrt\rho\to0$$
which is a much cleaner proof.
