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Why is it possible to calculate multivariable limits using polar coordinates? Let's say I'm looking for some $\lim_{(x,y) \to (0,0)}$ and I'm substituting $x = r cos\theta$ and $y = rsin\theta$ so that I can look at $\lim_{r \to 0}$. Why can I do this? Am I not just looking at "straight lines" going to $(0,0)$ now? What about all the other possible sequences that converging in straight lines to (0,0)?

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  • $\begingroup$ You can do it because each function is a product of two continuous functions, namely $r$ and $\cos(\theta)$ and $\sin(\theta)$. $\endgroup$ – manooooh May 21 at 20:43
  • $\begingroup$ It's really nothing more than the fact that $(x,y)\to(0,0) \iff |(x,y)|\to |(0,0)|$. Note, however, that this will not work in the more general case of $(x,y)\to(a,b)\neq(0,0)$. $\endgroup$ – MPW May 21 at 21:02
  • $\begingroup$ I think the problems OP has are 1. the $\theta$ dependence and 2. that we are assuming that they lie of the same circle, so that they're equidistant from the origin and moving in at the same rate. The former is obviously inconsequential here, but one might wonder about the latter. It turns out not to matter, but some care is needed. $\endgroup$ – cmk May 21 at 21:06
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While $r$ is going to $0$, $\theta$ is arbitrary. So, $\theta$ can freely change however it wants, as long as the radius is going to zero (that is, the convergence is uniform in $\theta$).

EDIT: See the following link for rigorous details: Polar coordinates for the evaluating limits

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  • $\begingroup$ I understand that $\theta$ is arbitrary but by declaring $x = cos\theta$ and $y = sin\theta$ am I not choosing one fixed $\theta$ for each sequence? Of course all $\theta$ will form a circle. But each sequence will be a straight line, won't it? $\endgroup$ – thanc May 21 at 20:48
  • $\begingroup$ Yes, I just realized exactly what you meant; I'll edit it. Still, we're only sending $r$ to $0$, so $\theta$ can be doing whatever it wants. $\endgroup$ – cmk May 21 at 20:50
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    $\begingroup$ Here's a post I just found with some rigor, if you like: math.stackexchange.com/questions/3131953/… $\endgroup$ – cmk May 21 at 20:56
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You are sending $x$ and $y$ to zero. This is satisfied by sending $r$ to zero because both $x$ and $y$ are proportional to $r$ in polar coordinates. Regardless of $\theta$, $\cos(\theta)$ and $\sin(\theta)$ are finite numbers, and thus as $r\rightarrow 0$, both $x=r\cos(\theta)\rightarrow0$ and $y=r\sin(\theta)\rightarrow0$. In this case you've turned a multivariable limit into a single variable limit, which is incredibly valuable.

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"Why is it possible to calculate multivariable limits using polar coordinates? Let's say I'm looking for some lim (x,y)→(0,0) and I'm substituting x=rcosθ and y=rsinθ so that I can look at lim r→0. Why can I do this? Am I not just looking at "straight lines" going to (0,0) now? What about all the other possible sequences that converging in straight lines to (0,0)?"

Yes, doing it that way is wrong! But if you show that the function goes to 0 as r goes to 0 without any reference to $\theta$, you are not taking the limit along any specific line. You are just saying that, for a point, (x, y), close enough to (0,0) (and in polar coordinates, the distance to (0, 0) is measured by r alone) the function is close enough to the limit.

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