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 $\rho\to0$.


This is where I get stuck. Thank you in advance for your help!

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

2 Answers 2


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


When $\sin^2(\theta)\ge\rho$ we have


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


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.

  • $\begingroup$ Can you please help me understand if I can reach that same result in polar coordinates? $\endgroup$
    – unnikked
    Aug 9, 2019 at 19:53
  • 1
    $\begingroup$ I think using polar coordinates is not the best way. $\endgroup$ Aug 9, 2019 at 19:54
  • $\begingroup$ The solution using this theory is deleted. $\endgroup$ Aug 9, 2019 at 19:55
  • 2
    $\begingroup$ What do you mean by your last comment? $\endgroup$ Aug 10, 2019 at 0:40

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