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Limits and polar co-ordinates

  1. Aug 20, 2006 #1
    Hi everybody,

    I have a question concerning limits of 2-variable real functions and complex functions of 1 complex variable. When we have a function f:A->R, (A<=RxR and (0,0)εΑ) and we are looking if the limit of f at (0,0) exists, then we can do a substitution to polar co-ordinates like this: (X,Y)->(Rcost, Rsint) and look for the limit of f(Rcost,Rsint) when R->0. If the limit doesn't exist or exists, but is a function of t, then we say that the limit of f(X,Y) when (X,Y)->0 doesn't exist. If the f(Rcost, Rsint), R->0 exists and is a constant, then it may be the limit of f(X,Y), but we have to check it by definition.

    I think all the above are correct, so my question is, why do we have to check by definition the limit in the last case? I have found examples that the limit doesn't exist, although we can find a constant one with polar co-ordinates, but I want a more precise explanation than examples.

    I have read that, by using this change of variables, we actually are looking for the limit in the straight lines Y(X)=Xtant , which seems quite sensible, but on the other hand, a mathematician who probably has misunderstood this specific topic, told me that if we find a constant limit, then it is correct, beacuse every point of the plain can be represented by polar co-ordinates, so depending on t, we can reach (0,0) through any line. And the "problem" was that I was not sure what to tell him to prove him wrong...

    (All the above apply to 1-complex-variable functions, too)

    To end up, I don't think i have a good grasp of the subject so any explanation would help. Sorry for the size of the post.

  2. jcsd
  3. Aug 20, 2006 #2


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    You're both right, I think. There's an ambiguity in notation here that I had never noticed before.

    There are two different ways of thinking about the expression:

    \lim_{r \rightarrow 0} f(r, \theta)

    The first way is what you're thinking -- that this is a limit parametrized by [itex]\theta[/itex]. You're thinking that this pointwise defines a function:

    g(\theta) := \lim_{r \rightarrow 0} f(r, \theta)

    The second way is what your mathematician friend (and I!) might be thinking: this is still a two-dimensional limit, so we're interested in whether it converges uniformly for all values of [itex]\theta[/itex].

    (edit: the LaTeX parser doesn't seem to be up and working. This should be readable eventually... otherwise click on the images to see what I was trying to write)
  4. Aug 20, 2006 #3
    I am not sure i got it (my bad English...), so do u say that if the limit is a constant when R->0, then the limit of f(X,Y) exists and is the above??
  5. Aug 20, 2006 #4


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    Given that I interpreted it to mean a two-dimensional limit and not a parametrized limit, yes. I.E. I took it to mean:

    For all epsilon, there exists a delta such that for any r and theta:
    0 < r < delta ==> |L - f(r, theta)| < epsilon
    Last edited: Aug 20, 2006
  6. Aug 20, 2006 #5
    Yes, but this f isn't the same as the initial one, because when we do the substitution we get f(Rsint,Rcost)=H(R,t) which is a different function. So, that doesn't mean that the limit f(X,Y) that we are looking for exists. But I am not sure why...
  7. Aug 20, 2006 #6
    Just to add the example i mentioned:

    f(x,y)={(x^2) e^[-(x^4)/(y^2)]}/y, limit? when (x,y)->(0,0)

    ((0,0) isn't in the function's domain but we can look for the limit)

    (Sorry for the notation but i think that there's a problem with Latex)

    So, using the polar co-ordinates we get that lim f(rsint,rcost)=0 when r->0.
    But, if we try to find the limit on the line h(t)=(t,t^2) we get lim...=1/e.
    Thus, the limit doesn't exist.

    I only mention the example for the ones that are not sure about the right/wrong of my question...
    Last edited: Aug 21, 2006
  8. Aug 21, 2006 #7
    How did you come up with that limit? If you plug in for (x,y) the line (t,t^2) and you let t ->0 you get [itex]\lim_{t\rightarrow 0} f(t,t^2)=\lim_{t\rightarrow 0} e^{-1}\cdot t^4[/itex], right?

    The only thing you have to treat carefully in this context is derivatives. Let's say you wanted to know the derivative of f at (x,y). Can you imagine why it would be false to express f in terms of polar coordinates and then take the derivative in terms of r?
    Best regards...Cliowa
  9. Aug 21, 2006 #8
    I don't think you get a "different" function in a fundamental sense. The behaviour of this function surely is exactly the same, the only thing you changed are the coordinates, so you must express your desires in terms of the new coordinates. But clearly (x,y)->0 is equivalent to polar r->0. Do you see what I'm saying?

    One could also decompose H(r,t) into two separate functions:
    [itex]H(r,t)=f(x,y)\circ P(r,t)[/itex], where P(r,t)=(r cos(t), r sin(t)). So P translates the coordinates from polar to the cartesian ones. Now you can look at the limit of the composition.
  10. Aug 21, 2006 #9
    Sorry about that. I gave you the wrong function by mistake. Check my previous post again. I have edited it
  11. Aug 21, 2006 #10
    That is exactly my question. Do you think of it with a geometrical aproach, too? I mean, I haven't really used polar co-ordinates, and generally I have some difficulty with anything other than cartesians, but this initial transformation (x,y)->(rcost,rsint), if not seen geometricaly, gives a function that is the composition of f and the above one.

    Actually, the problem is, I am not sure that I totally get what you mean by saying that clearly cartesian (x,y)->(0,0) is equivalent to polar r->0. If by "polar" you mean the change of variables that I suggested to find a possible limit, then you are wrong, and the example shows that. If you mean something else, I would like you to explain it to me a bit.

  12. Aug 29, 2006 #11
    Yes, I do think of it with a geometrical approach. The polar coordinates are just different coordinates for the same space [itex]\mathbb{R}^2[/itex].

    By polar I mean the change of coordinates discussed above. Let's check my equivalence:
    1) If (x,y)->(0,0) in cartesian coordinates this implies that in polar coordinates r->0, as you can't find a value of t such that sin(t)=cos(t)=0. Ok?
    2) If r->0 in polar coordinates, clearly also (x,y)->(0,0) in cartesian coordinates.
    So what I wrote was absolutely correct.
    The problem with your example is this: If it is non-sensical to have a limit in cartesian coordinates, it is quite likely to be non-sensical to have one in polar coordinates. And that's the case here, as your function is not defined on the whole x-y-plane. One could move along the x-axis (x,0) with x->0 and all along the way the function value would be undefined. So it does not make any sense at all to ask for a limit when (x,y)->(0,0).
    Why don't you try the limit for (x,y)->(0,1)?
    best regards...Cliowa
  13. Aug 29, 2006 #12
    First of all, thanks for your answer.
    If it is correct, then I think that we could prove that the limit doesn't exist, using only polar co-ordinates. And I will explain why this limit has sense in my opinion.

    The function I mentioned is defined on RxR-{(x,0), x: real}. The point (0,0) is a limit point of the domain of the function so I think that we can look for a limit at (0,0).

    That's all! I am still not sure of the right answer, so i will add nothing more for now.
  14. Aug 29, 2006 #13
    Still, there is no vicinity of the point (x,y)=(0,0) such that the function is well-defined in the whole neighbourhood. That's the problem. Remember that the criterion for a functional limit (continuity) is this: For any (!) sequence [itex]a_k=(x_k,y_k)[/itex] with [itex]a_k=(x_k,y_k)\rightarrow (0,0) [/itex] as [itex]k\rightarrow\infty[/itex] it is necessary (and sufficient for continuity) that [itex]f(x_k,y_k)\rightarrow a\in\mathbb{R}, k\rightarrow\infty[/itex]. Clearly, this does NOT hold in your situation.

    You could however discuss "directional limits", i.e. confine your function to a line (which is not the x-axis of course) in [itex]\mathbb{R}^2[/itex]. This could easily be done using polar coordinates (just fix an angle [itex]t\neq k\cdot\pi, k\in\mathbb{Z}[/itex] and let r->0).

    I hope I'm on the right way convincing you. Best regards...Cliowa
  15. Aug 29, 2006 #14
    I think that the problem may occur because of various definitions of things but in the case of the limit of the function, I believe (and hope) that there is a universal one... I have checked the definition of limit in three books (Calculus 1,2 and Complex Analysis) and in all of them the definition of limit goes like that. If a is a limit point of a function f defined in domain D and for every ε>0 there exists δ>0 so that if 0<|x-a|<δ, with xεD, then |f(x)-k|<ε , then we say that limf(x)=k, x->a.

    Tha same goes for the criterion with the sequences: for any sequence {a(n)} with a(n)εD.....

    I am not sure which is the most widely accepted definition, because, for example, in mathworld there is no mention in the part (..."xεD"...), but i think that we can still talk about limits (and not directional ones) even in points where no neighbourhood of them is a subset of the domain of the function
  16. Aug 29, 2006 #15
    Ok, now this is getting a small definition-battle (which neither one of us might win), but I still think that even though your textbooks define limits that way it is non-sensical to think about a limit in our case (and cases alike). Here's why: One of the main reasons (if not the main reason) for discussing limits of functions at the border of their domains is continuity. The question is this can one extend the domain in such a way, that the resulting function is continuous? This could be done by assigning a special value to the newly added point of the domain, for example the limit (if there is one and only one). In fact, this is a definition widely used, I think. And it's a sensible one.

    In the case of the function you presented there is no [itex]a \in \mathbb{R}[/itex] such that the function [itex]F:\begin{cases}\frac{(x^2)}{y} e^{-\frac{(x^4)}{(y^2)}}, &(x,y)\in\mathbb{R}^2 \setminus \{(x,0);x\in\mathbb{R}\}\\
    a, &(x,y)=(0,0)\end{cases}[/itex] is continuous. Therefore the limit does not make sense, in my opinion. If you take a certain value for a, there is no reason why this should be considered the limit, there's no reason for this to be declared any "better fitting" than others. You can't argue for it (at least I couldn't). Do you get my point?
  17. Aug 30, 2006 #16
    I think we have done a circle here and reached my initial question. I have said that the above limit doesn't exist, and I get that you agree with that, but by using polar co-ordinates we get that it does exist and is equal to 0. And that was my question, why polar co-ordinates don't prove that the limit exists but only show what the possible limit, if it exists, may be. If they were to be equivalent, then this just shouldn't happen.
  18. Aug 30, 2006 #17
    I think I argued that the limit is non-sensical in general, and so it is using polar coordinates. I tried to show you that there is a problem with the domain of the function. Polar coordinates do adress that issue, as for some values of t the resulting function you get when taking the limit does not make any sense. But this is a rather general feature of mathematics: You can't simply plug in some stuff and then switch off your brain expecting the algebra to produce the right result. Please don't take this as an insult, I only mean to express that the formalism won't solve everything.

    I don't wanna say again that they're equivalent, but the point is this: If you take the algebraic expression for your function in polar coordinates, then take the limit for r->0 and look what you get this does not mean you found a limit: You can't ignore the domain issue. That's something you need to consider too. So when you substitute polar coordinates it may seem that you do get a limit, but the limit of a function isn't simply the limit of the algebraic expression. There's more to it (at least in the classical sense). Wouldn't you agree?
  19. Aug 30, 2006 #18


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    Yes, but the example you give is not defined for any (x, 0), [itex]x\ne 0[/itex]. Of course, converting to polar coordinates, you would find that for [itex]\theta= \frac{\pi}{2}[/itex] or [itex]\theta= -\frac{\pi}{2}[/itex] you get no limit and so conclude that the original function has no limit at (0,0).

    If you changed to definition to
    [tex]F:\begin{cases}\frac{(x^2)}{y} e^{-\frac{(x^4)}{(y^2)}}, &(x,y)\in\mathbb{R}^2 \setminus \{(0,0)\}\\
    a, &(x,y)=(0,0)\end{cases}[/tex]
    The limit at (0,0) would exist and be 0.
    Last edited by a moderator: Aug 30, 2006
  20. Aug 30, 2006 #19
    Are you kidding me? Do you really want to say that a=0 in what I wrote? If you look along the parabola h(t)=(t^2,t) (as c0nfused suggested) you get 1/e as the "limit". Right?

    And HallsofIvy, btw, the function you "defined" is ill-defined. What's the point of that definition?
    Last edited: Aug 30, 2006
  21. Aug 31, 2006 #20


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    No, I didn't say that a= 0 in what you wrote. I clearly said that what you wrote is not defined for (x, 0) and so could not possibly have a limit at (0,0) even if you write it in polar coordinates. I concede that what I wrote is also not defined for such points. I'm not sure I would say that my function is any more "ill-defined" than yours (yes, I just didn't notice the denominators of y) since it is standard to limit the domain even when it is not explicitely given. In fact the function I defined is exactly the same as yours so does not have a limit at (0,0).
    I thought your point was that changing to polar coordinates would give an incorrect limit. My point was that since your function is not defined for (x, 0), polar coordinates does not give a limit either.

    (And you mean on the parabola (t, t2), don't you?)
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