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Proving Limits

  1. May 22, 2009 #1
    1. The problem statement, all variables and given/known data

    http://img40.imageshack.us/img40/39/20688555.gif [Broken]

    a) Show that if C is any straight line through (0,0) then [tex]\lim_{(x,y) \to (0,0)}[/tex] along C exists and equals 1.

    b) Show that the limit as (x,y) -> 0 doesn't exist.


    2. Relevant equations



    3. The attempt at a solution

    I really need help with this question! I know that x2 is a parabola and any straight line through the origin either intersects the parabola at some point and remains above it until 0 is reached (or lies on y=0, in which case [tex]y \leq 0[/tex] and [tex]f(x,y) = 1[/tex]) but still I don't know how to "prove" part a).
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 22, 2009 #2

    tiny-tim

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    But you've done it! :smile:

    If, sufficiently close to the origin, it remains above the parabola until 0 is reached, then the value along that part of the line is 1, so the limit exists …

    what's worrying you about that? :confused:
     
  4. May 22, 2009 #3
    Hi!

    In part b) asks to 'show' that the limit as (x,y) -> 0 doesn't exist...
     
  5. May 22, 2009 #4

    tiny-tim

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    Well, it obviously doesn't exist …

    to prove it, use the definition of limit (y'know, the epsilon and delta thing) :wink:
     
  6. May 22, 2009 #5
    Yes I see. I know how the epsilon & delta proof works but I don't know how to exactly apply it to this problem. I appreciate it if you could help me get started on this, thanks.
     
  7. May 23, 2009 #6

    tiny-tim

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    Well, if the limit exists, and is 1, say, then that means that given any epsilon > 0, we can find a ∂ > 0 such that √(x2 + y2) < ∂ means that |f(x,y) - 1| < epsilon.

    So ty it with epsilon = 1/2 :wink:
     
  8. May 23, 2009 #7

    HallsofIvy

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    Simpler, I think: try approaching (0,0) along the curve y= (1/2)x2. In that case, y is always less than x2 so f(x,y)= 0.
     
  9. May 27, 2009 #8
    You mean [tex]\epsilon = \frac{\delta}{2}[/tex]?
     
  10. May 27, 2009 #9

    Mark44

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    No, he meant epsilon = 1/2.
     
  11. May 27, 2009 #10

    HallsofIvy

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    Using an [itex]\epsilon- \delta[/itex] proof, [itex]\epsilon[/itex] is "given" and [itex]\delta[/itex] depends on [itex]\epsilon[/itex], not the other way around.

    But I still think my suggestion is simpler. I'll just go off and sulk!
     
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