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F(x) integral converge, f(x) uniformly continuous ,prove that f(x) limit = 0

  1. Dec 14, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\int_{a}^\infty\ f(x) dx [/tex] <--- converge
    f(x) uniformly continuous in [a,[tex]\infty[/tex]]
    prove that [tex]lim_{x\rightarrow \infty} f(x) = 0[/tex]

    2. Relevant equations

    3. The attempt at a solution

    I know that if f(X) has a limit in [tex]\infty[/tex] it has to be 0
    I think that the solution has to be conected to the fact that if f(x) uniformly continuous ,there is a M that |f'(x)|<M,
    I think I can prove that if f(x) does not have limit it's Derivative has to change infinite times form + to - , so f(x) has to go up and down infinite times..
    and when it go up there is a limit to how low her max can be,
    I think I have to put it all together with Cauchy test
    But I cant seem to do it.
    Thank you
  2. jcsd
  3. Dec 14, 2009 #2


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    Homework Helper

    Try a proof by contradiction. If f(x) does NOT converge to zero, then there is an epsilon>0 such that for all N>0 there is an x>N such that |f(x)|>epsilon. Do you agree with that statement? If you draw a picture, then that means that the graph of f(x) must have an infinite number of points where |f(x)|>epsilon stretching off to infinity. If f(x) is also uniformly continuous, then there is a corresponding delta such that |f(x)-f(y)|<epsilon/2 for |x-y|<delta. That means all of those x's in your picture correspond to a bump in the graph of f(x) with area greater than delta*epsilon/2. Can you show this contradicts the convergence of the integral?
  4. Dec 15, 2009 #3
    Thank you very much
    That was very helpful
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