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Boundary points and limit of f(x,y)

  1. Sep 25, 2014 #1
    Let f(x,y) be defined by

    f(x,y) = [x2y2]/[x2y2 + (x-y)2]
    a) Find the domain of the function f.
    b) show that (0,0) is a boundary point of the domain of f
    c) Compute the following limit if it exists:
    lim (x,y) ---> (0,0) f(x,y)

    3. The attempt at a solution
    a) I first change the value (x-y)2 to (x2+y2-2xy). I then looked at it logically and said that the denominator obviously can't be equal to zero. And if I rewrote it to look like
    x2y2+x2+y2-2xy, I can see that 2xy is the only term that is not squared, which means it is the only term capable of being negative. So I wrote that x2y2+x2+y2 > 2xy
    That is it for the domain. Is my logic correct here?

    b) I believe I somewhat have an understanding of boundary points. Basically it means that for a neighborhood of x (an open set, U, that x is contained in), at least one point of U will be contained inside and outside of A. Correct? This being said, I'm still not really sure how to do this part of the problem. The domain at (0,0) is undefined because x2y2+x2+y2 = 2xy, causing you to divide by zero. Is this in a sense proving already that it is a boundary point? Because x=0 and y=0 by themselves are contained in the domain but when both are, they are not. Am I completely wrong here in that assumption?

    c) I'm really not sure how to go about trying to calculate this limit. Limits are not my specialty. I can't think of a path that doesn't lead to 0, but that isn't proof enough to say that the limit definitively exists.

    Some help would be greatly appreciated! Thank you!
     
  2. jcsd
  3. Sep 25, 2014 #2

    LCKurtz

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    No. The only thing that can go wrong is for the denominator to equal zero. There is nothing wrong with negative values of ##f##.

    You have to show that for any point not in the domain, there is a point arbitrarily close to it that is in the domain.

    If you suspect the limit doesn't exist, see if you can find two paths that give different results. If you can, you have shown the limit doesn't exist. Try that first.
     
  4. Sep 25, 2014 #3
    I tried looking up the domain and limit of the function via wolfram alpha. It gave me that my domain was correct, so I'm not sure what to do about that. I still don't know what to do about the boundary point either. And I suspected that the limit DOES exist. And Wolfram alpha again put that the limit was equal to zero. I have no idea how to prove if a limit does exist. Only how to prove that it doesn't. I'm still completely lost with this one.
     
  5. Sep 26, 2014 #4

    LCKurtz

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    Well, you have two choices. You can blindly type stuff into Wolfram or you can think about what I told you and address it. What did I tell you about the domain? What is your reaction to that? What paths have you actually tried to see if you get different answers? Show me something.
     
  6. Sep 26, 2014 #5

    haruspex

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    LCKurtz is right, only the places where the denominator is zero are outside the domain. Check you are interpreting the Wolfram answer correctly.
    A useful technique for such problems is to consider approaching the singularity along a straight line, in this case y = ax or the special case x = 0. Then consider whether all values of a give the same limit.
    If they do, to get a formal proof just apply the usual epsilon-delta logic: given epsilon > 0, find a delta such that if (x,y) is within delta of (0,0) then etc.
     
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