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Limits Question

  1. Apr 5, 2003 #1
    How can one prove that:

    lim (x,y)->(0,0)

    (x^4+y^4)
    ---------
    (x^2+y^2)

    = 0

    I keep getting 0/0 no matter what I do to the equation. Anyone have any pointers?
     
    Last edited by a moderator: Apr 5, 2003
  2. jcsd
  3. Apr 5, 2003 #2

    Hurkyl

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    Apply the epsilon-delta definition of a limit, or try a clever substitution.

    Hurkyl
     
  4. Apr 5, 2003 #3
    Thank you for the quick response. I'll go and try it out now.
     
  5. Apr 8, 2003 #4

    HallsofIvy

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    One of the difficulties with limits in more than one variable is that there are an infinite number of different ways to "approach" the target point. In order for the limit to exist, the result must be the same along any path.


    Since the "epsilon-delta" definition Hurkyl mentioned used the distance from the target point the best way to use it is to change to polar coordinates so that r measures the distance from from (0,0).

    Note that x= r cos([theta]) and y= r sin([theta]) so that
    x<sup>2</sup>+ y<sup>2</sup>= r<sup>2</sup> and
    x<sup>4</sup>+ y<sup>4</sup>= r<sup>4</sup>(cos<sup>4</sup>([theta])+sin<sup>4</sup>([theta]).

    That should make it easy.
     
  6. Apr 8, 2003 #5

    HallsofIvy

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    Oops, wrong forum, wrong symbols. Well, the math is still correct.
     
  7. Apr 8, 2003 #6
    Would using L'Hospital's rule (successive differentiations of numerator and denominator) be cheating?
     
  8. Apr 9, 2003 #7

    HallsofIvy

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    Since this is a function of two variables, HOW, exactly, would you apply L'Hospital's rule?
     
  9. Apr 9, 2003 #8
    Thanks HallsOfIvy. That was much easier to work with
     
  10. Apr 9, 2003 #9
    HallsofIvy
    (d2N/dxNdyN)((x^4+y^4)/(x^2+y^2)),

    where the derivatives are partial.
     
  11. Apr 11, 2003 #10
    the proof is as follows.

    lim(x,y)->(0,0)

    x^4 + y^4
    ---------
    x^2 + y^2

    =

    (x^2 + i*y^2)*(x^2 - i*y^2)
    ---------------------------
    x^2 + y^2

    =

    x^2 - i*y^2

    as x and y go to zero this value approches 0.

    Not that when they are zero the value of the function is NOT zero.
     
  12. Apr 11, 2003 #11
    ObsessiveMathsFreak
    Is that value then undefined?
     
  13. Apr 11, 2003 #12
    Why

    (x^2 + i*y^2)*(x^2 - i*y^2)
    ---------------------------
    x^2 + y^2

    =

    x^2 - i*y^2

    ?

    x^2 + i*y^2<>x^2 + y^2

    Try this way :

    (x^4+y^4)/(x^2+y^2)=((x^2+y^2)^2-2*x^2*y^2)/(x^2+y^2)=
    =1-2*x^2*y^2/(x^2+y^2);
    Now lim (x^2+y^2)-2*x^2*y^2/(x^2+y^2) = lim (x^2+y^2) -
    lim 2*x^2*y^2/(x^2+y^2)= 0 -1 / lim (x^2+y^2)/2*x^2*y^2=
    =-1 / lim (1/(2*x^2)+1/(2*y^2))=-1/infinity = 0;

    I hope this is correct...

    By the way...HallsofIvy...
    I don't think your "notation" is correct...
    Because if you say x=r*cost and y=r*sint you practically say
    x=k*y, which is not correct...x could be equal to y^2...
    (because x->0 and y->0 means that r->0, because cost and sint
    can't -> 0 in the same time)
    See ya...
     
    Last edited: Apr 11, 2003
  14. Apr 11, 2003 #13

    HallsofIvy

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    I SAID that was converting to polar coordinates. The point with coordinates r and theta in polar coordinates has x= r cos(t) and
    y= r sin(t) in cartesian coordinates. Believe it or not I am completely aware that as (x,y)-> (0,0), r-> 0! That was the whole point! Since r measures the distance from (0,0) to the point, the two variables x and y going to 0 reduces to the single variable r going to 0.
     
  15. Apr 11, 2003 #14
    (x,y) is a point ?
    Not a pair of variables ?
    My mistake then...
    Sorry...
    But there's no real need to consider them coord of a point...
     
  16. Apr 11, 2003 #15

    HallsofIvy

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    If you are willing to ASSUME that you can you can
    treat 1/(1/x^2+ 1/y^2) as x and y going to 0 as
    "1/(infinity+infinity)" and declare that "1/infinity" is 0, then there is no real need to be precise at all. Yes, your method does work (and is very clever) in this example but I suspect that most mathematics professors would want you to show a little more understanding of what you are doing.
     
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