# Limits are different when using rectangular/polar coordinates

As part of calculating another limit, I have the following limit

$$(x, y) \to (0,0) ~~~~~\frac{x}{x^2 + y^2}$$

When using rectangular coordinates, going through the path $$y = mx$$, we get

$$\frac{x}{x^2 + m^2x^2} = \frac{1}{x(1+m^2)} = \frac{1}{1 + m^2}\frac{1}{x}$$

So as we take the limit $$\lim x \to 0$$ we get $$\infty$$

In polar coordinates, however,

$$\frac{x}{x^2 + y^2} = \frac{rcos\theta}{r^2} = \frac{cos\theta}{r}$$

And as $$r \to 0$$ and $$\theta \to 0$$, we have the indeterminate form $$0/0$$ and can't use L'hôpital's rule because we have different variables on the numerator and denominator.

So the limit doesn't exist using polar, but it's $$\infty$$ using rectangular. Which is right?

• Why would it be $+\infty$ in rectangular? You only checked some lines through the origin, that's not enough! – Botond Sep 5 '19 at 21:41
• Taking the limit along paths $y=mx$ does not mean the limit exists in rectangular, even if they all approach the same thing. – Ninad Munshi Sep 5 '19 at 21:41
• path $y=mx$ corresponds to $\theta=\arctan(m)$, so $\sin\theta$ is a non-zero constant – J. W. Tanner Sep 5 '19 at 21:43
• by the way, did you mean $x=r\color{red}{\cos}\theta$? – J. W. Tanner Sep 5 '19 at 21:44
• @J.W.Tanner yes, thanks. – Sigma Sep 5 '19 at 21:48

You have NOT shown it's $$\infty$$ using rectangular coordinates.

You've only shown that's the case for your specific path of approach ($$y=mx$$). It's possible you could get a different result for a different path of approach.

To make that conclusion, you would have to show that the limit is the same for every path of approach.

The polar coordinate condition corresponding to the rectangular coordinate path $$y=mx$$ as $$x\to0$$ (where we're taking the limit of a constant times $$1/x$$) would be $$\theta=\arctan(m)$$ as $$r\to0$$. In this case, we're taking the limit of $$\cos\theta/r$$ as $$r\to0,$$ and $$\cos\theta\ne0$$, so it's just like the rectangular case.

Apart from the fact that you have no reason in the rectangular calculation to conclude that the limit is $$\infty$$ (by the way the limit of $$1/x$$ at $$0$$ is $$\pm\infty$$ depending on your approach, but I digress), to calculate the limiting value of $$\frac{\cos\phi}{r}$$ at the origin in polar coordinates, you do not let $$\phi\to 0,$$ otherwise, again, you'd be focusing only on one particular direction (even if you did, you'd not get the form $$0/0,$$ as claimed but $$1/0=\pm\infty,$$ as before, but again this is tangential to the main point).

We let $$\phi$$ be indeterminate (any angle) as we let the radius $$r$$ shrink to nothing. Of course, we must make sure that whatever function of $$\phi$$ we have is always defined and bounded for all $$\phi,$$ which in this case it is. Also, for this case, we have to make sure the numerator does not vanish. But it does whenever $$\phi=(2k+1)\fracπ2,$$ so we need to question whether the limit does exist.

• To be fair, all you've shown is that the limit is bounded from above by $\infty$, which isn't saying much. – Ninad Munshi Sep 5 '19 at 22:09
• @NinadMunshi I went through my answer again and discovered you were probably referring to the latter part, which wasn't the maim point, but which indeed doesn't show the limit is infinite. But the polar calculation above clearly shows that. I'll repair the rectangular forthwith. – Allawonder Sep 5 '19 at 22:19
• If $\phi = \pi/2$ what is the limit? The limit in such functions is defined if all paths agree on the value of the limit. – Niki Di Giano Sep 5 '19 at 22:32
• @NikiDiGiano Wow, I'd overlooked that. Thanks for pointing it out. Answer adjusted. – Allawonder Sep 5 '19 at 22:58
• "We need to make sure the numerator does not vanish" and "We need to question whether the limit does exist." What? The limit is $\infty$ for all $\phi \neq (2n+1)\pi/2$ and $0$ everywhere else as far as paths are concerned. This is enough to show that the limit of $f(x, y)$ does not, in fact, exist. – Niki Di Giano Sep 5 '19 at 23:01