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Limit of multivarible functions

  1. Apr 6, 2009 #1
    Let's say a two-variable function f(x,y), consider the limit at (x,y)=(a,b).
    If for any path y=h(x) approaching (a,b), the single variable functions f(x,h(x)) have the same limit, can I say that the limit of f(x,y) at (a,b) exist(using epsilon-delta definition),and how to prove?
    Thanks.
     
  2. jcsd
  3. Apr 6, 2009 #2

    CompuChip

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    I am quite confident that the statement is true. You can probably even prove it. Let [itex]\epsilon > 0[/itex].
    Then you know that for any path y = h(x) there exists a [itex]\delta = \delta_\epsilon(h)[/itex] such that
    [tex]|| (x, y) - (a, b) || < \delta \implies |f(x, h(x)) - L| < \epsilon,[/tex]
    where L is the supposed limit of f(x, y) at (a, b). You want to show that there exists a "global" [itex]\delta_\epsilon[/itex] such that
    [tex]|| (x, y) - (a, b) || < \delta \implies |f(x, y) - L| < \epsilon[/tex]
    for any y. On a nice neighbourhood of (a, b) you can always find a path h from (x, y) to (a, b) and you have the corresponding delta(h). So you could think of something like taking some minimum (infimum) of all those delta(h)'s. The real "challenge" of the proof, would then be to show that this infimum is non-zero (i.e. you can really find a delta strictly > 0 to satisfy the definition).
     
  4. Apr 6, 2009 #3
    \delta = \delta_\epsilon(h)
    what does it mean? Functional?
     
  5. Apr 6, 2009 #4

    CompuChip

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    I meant to say that the delta that you need will depend both on epsilon and on the path. I.e. different paths may need different delta's.
    So strictly speaking, yes, it is a functional (for fixed epsilon, it assigns to any given path a number), but you don't really need to view it that way. It was just to prevent confusion with "the" delta you use in the multi-variable limit.
     
  6. Apr 6, 2009 #5
    well, thanks a lot, I'll try
     
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