I was trying to calculate the following limit: $$\lim_{(x,y)\to(0,0)} \frac{xy}{x-y}$$

I used polar coordinates: $x=r\cos\theta$ and $y=r\sin\theta$. But this gives me: $$\lim_{r\to 0} \frac{r\cos\theta\cdot r\sin\theta}{r\cos\theta-r\sin\theta}=\lim_{r \to 0} r\frac{\cos\theta\sin\theta}{\cos\theta-\sin\theta}=0$$

But Wolfram Alpha says that this limit does not exist. What is the problem in what I did? Is it because of the fact that if $\theta = \frac{\pi}{4}$ I am dividing by $0$? (My guess is that it is in fact the problem and by applying L'Hôpital's rule I will find out the limit does not exist.)

  • 1
    $\begingroup$ The proof would work if the factor in terms of $\theta$ were bounded by some constant. But it's not bounded, so no matter how small $r$ is, we can find a value of $\theta$ where $\cos \theta \sin \theta / (\cos \theta - \sin \theta) > 1/r^2$... $\endgroup$
    – aschepler
    Mar 8, 2021 at 22:43

1 Answer 1


The first problem is that $\frac{xy}{x - y}$ doesn't even make sense if $x = y$, because the denominator is then zero. But even if we exclude the line $x = y$ from the domain, the limit still doesn't exist. Essentially, the problem is that we can approach $(0, 0)$ along a path that approaches the line $x = y$ rapidly enough that $x - y$ approaches zero faster than $xy$, forcing the limit along this path to be nonzero. (For example, $y = x^3 + x$ gives such a path.) On the other hand, the function is constant zero along the line $x = 0$, so if the limit exists, then it has to be zero. Thus the limit can't exist.

Rewriting it in polar coordinates doesn't change this, and the same phenomenon still occurs with that term involving cosines and sines of the angle.


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