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A limit problem

  1. Jun 2, 2003 #1
    [SOLVED] A limit problem

    I just need some help showing how this limit systematically follows from the limit rules:

    Code (Text):

               # x     2
          1   #       t           1
    lim  ---  #  ---------- dt = ---
    x->0   3  #     2             3
          x  # 0   t  + 1
     
    My first chain of thought led to breaking the expression up as follows:
    1/x^2 * ( 1/x * Integral[0..x, t^2/(t^2+1)] )

    Then I just kind of figured that the subexpression on the right was the average value of the function being integrated from 0..x and as x->0 the average value would approach x^2/(x^2+1), which led to:

    1/x^2 * x^2/(x^2+1) = 1/(x^2+1)
    which would approach one as x approached zero.

    But clearly that's wrong (not surprisingly since I made a sketchy move in the middle) since the answer is one-third. Can anyone show me how to do this?

    edit: possibly w/o actually integrating because this is an exercise in which you're expected to know the FTofC but not how to integrate that.

    edit2: oh, not supposed to no l'hospital's rule either.
     
    Last edited by a moderator: Jun 2, 2003
  2. jcsd
  3. Jun 2, 2003 #2

    Hurkyl

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    Your first thought would have worked if you only had 1/x out front instead of 1/x3. (because your limit would essentially be a derivative)

    So the trick is to rewrite it in a form in which the fundamental theorem of calculus applies! In particular, if you can do a substitution in the integral so the bounds of integration are from 0 to x3, then you can use your thought to evaluate the limit.
     
  4. Jun 2, 2003 #3
    Okay, I think this works then.

    I tried to change Integral[0..x, t2/(t2+1)] into Integral[0..x3, f(t)]

    Integral[0..x, t2/(t2+1)] = Integral[0..x3, f(t)]
    d/dx[ Integral[0..x, t2/(t2+1)] ] = d/dx[ Integral[0..x3, f(t)] ]
    x2/(x2 + 1) = f(x3).3x2
    f(x3) = 1/(3(x2+1)), (x != 0)
    f(x) = 1/(3(x2/3+1))

    so..
    Limit[ x->0, 1/x3 Integral[0..x3, 1/(3(t2/3+1))] ]

    Which is the average value of the function inside the integral from 0..x3, which approaches f(0) as x approaches zero, which would be 1/3. I hope that's a sufficient way to solve the problem.
     
    Last edited by a moderator: Jun 2, 2003
  5. Jun 2, 2003 #4

    Hurkyl

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    That's an interesting way to change the limits! But you got the right answer in the end, I'm gonna have to look at it and see why it works; I never thought to do it that way.

    Incidentally, I was thinking doing the substitution t = s1/3. t = x => s = x3, t = 0 => s = 0, and the integral became

    ∫0..x3 s2/3 / (s2/3 + 1) * (1/3) s-2/3 ds

    which is precisely the integral you got.


    Anyways, then you're left with the form:

    L = limx->0 1/x3 &int0..x3 g(s) ds

    We are also permitted to substitute in the limit variable, and I will do so to make things simpler. x3 = y

    L = limy->0 1/y &int0..y g(s) ds

    By the fundamental theorem of calculus, if G(s) is the antiderivative of g(s):

    L = limy->0 (G(y) - G(0)) / y = G'(0) = g(0)

    So that's how you rigorously justify your last step.
     
  6. Jun 2, 2003 #5
    Thanks!
     
  7. Jun 2, 2003 #6

    Hurkyl

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    I guess, for the sake of completeness, I should specify that when I changed the limit variable, I had to use a function that is continuous and invertible near x = 0 (I think that alone is sufficient to permit the operation). y(x) = x1/3 satisfies that condition.
     
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