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Finding limits in 3D. How do you know on what line to approach?

  1. Mar 20, 2013 #1
    So say I have an arbitrary function and I want to know it's limit as x,y approaches 0.

    I could test what happens when the x axis approaches 0, y axis as it approaches 0 but there are some functions where I'm told that I also need to test what happens when y=mx approaches 0, and then y=x^2 and x=y^2 and I'm quite confused as to how we know what lines to approach 0 on and what functions those lines are in 3D and how ones works that out. I've basically lost any form of intuition here.
  2. jcsd
  3. Mar 20, 2013 #2
    Those are only pathological examples of various ways that a function can fail to have a limit at a point, so that one doesn't labor under the false impression that one need only test along all lines, or a selection of curves through the point in question. Ie., they tell you why you should not take the approaches in question at face value. You must apply the full definition of the limit in more than one dimension to prove that a given function f(x, y) has a limit L at a point (a, b) and show that all the values of f in a disk of radius ε about (a, b) approach L as ε approaches 0. You cannot take a shortcut and simply write y as a function of x for any arbitrary function unless you show some other reason why this technique will give the correct limit L.
  4. Mar 20, 2013 #3


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    I would not call them "pathological". In a very specific sense "almost all" function will give the same limit along all straight lines but NOT have a limit, since you can get different results by approaching a point along a curve.

    Perhaps the simples thing to do in three dimensions (or two dimensions) is to change to spherical (polar) coordinates so that as long as the limit as [itex]\rho[/itex] (r) goes to 0 exists, independent of the other coordinates, the limit will exist.
  5. Mar 20, 2013 #4
    I just mean it seemed arbitrary to me that they would substitute in these random functions in the textbook and I was trying to understand what the reasoning was behind deciding to substitute in something like y=x^2...
  6. Mar 20, 2013 #5
    I really like this idea, that going along a curve gives you a different limit

    The one we always were taught was (xy^2)/(x^2 + y^4), the contour plot:

    http://www5a.wolframalpha.com/Calculate/MSP/MSP46721d7g190ff4673c600001df0fie0bd6g5dbg?MSPStoreType=image/gif&s=19&w=299.&h=300.&cdf=RangeControl [Broken]

    The limit at the origin is 0 along any straight line, yet there are two contours x = y^2 and x = -y^2 which are 1/2 and -1/2 everywhere, including through the origin!

    This is a way of demonstrating that a constant limit along every straight line is not sufficient to prove the limit exists. The parabolic curves are counterexamples that are sufficient to prove that the limit doesn't exist, so that's why they're picked. In the above example, rather than being randomly chosen, these parabolas are fairly evident from the contour plot.
    Last edited by a moderator: May 6, 2017
  7. Mar 20, 2013 #6
    They designed the function to have a particular limit at that point that was different from limits along straight lines through the point in order to illustrate the fact that you shouldn't assume taking the limit along all straight lines through the point would give you the correct limit. It was arbitrary; they could have designed their function to behave in many other ways to illustrate the same point with a different curve.
    For example, here's a designer function where the design is much more obvious:
    f(x, y) = \begin{cases}1, & y = e^x\\ 0, & y \neq e^x\end{cases}
    If we attempt to approach the point (0, 1) along any straight line or polynomial curve, we get 0 as the limit of f, but approaching along [itex]y = e^x[/itex] gives a different limit, and thus the limit of f at (0, 1) does not exist. There are many ways to hide this behavior so that it is not as obvious which curves will give different limits, and thus help caution the student about limits that may seem obvious.
    Last edited: Mar 20, 2013
  8. Mar 30, 2013 #7
    Is it possible to use polar co-ordinates to test for limits when your not checking the limit as f approaches (0,0), e.g. we want to see the limit as f approaches (1,0)
  9. Mar 31, 2013 #8
    I'd have thought you can shift the function to (0,0) by replacing x with x-1, and then convert to polars.


    lim(f(x,y)) at (1,0) = lim(f(x-1,y) at (0,0)
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