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?
 A: 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.
A: 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.
A: 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.
