# $\lim_{(x,y)\to(0,0)} \frac{xy}{x-y}$

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.)

• 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 $\cos \theta \sin \theta / (\cos \theta - \sin \theta) > 1/r^2$... Commented Mar 8, 2021 at 22:43

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.