# 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!

• Compare with this or possibly this. Using polar coordinates works on many occasions, but sometimes it leads to a mess. Aug 9, 2019 at 19:52

## 2 Answers

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.

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.

• Can you please help me understand if I can reach that same result in polar coordinates? Aug 9, 2019 at 19:53
• I think using polar coordinates is not the best way. Aug 9, 2019 at 19:54
• The solution using this theory is deleted. Aug 9, 2019 at 19:55
• What do you mean by your last comment? Aug 10, 2019 at 0:40