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Math Help - Urgent and important??! integral inequality

  1. #1
    Junior Member miss_lolitta's Avatar
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    Urgent and important??! integral inequality

    Hi guys,

    please can someone here help me to prove that:

    Assume that:

    \int_1^{\infty}c(x)^{1/2}<\infty
    this implies that:

    \lim_{x \to\infty}{x c(x)^{1/2}}=0


    thanks in advance
    Last edited by ThePerfectHacker; December 17th 2006 at 01:29 PM.
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    Quote Originally Posted by miss_lolitta View Post
    Hi guys,

    please can someone here help me to prove that:

    Assume that:

    \int_1^{\infty}c(x)^{1/2}<\infty
    this implies that:

    \lim_{x \to\infty}{x c(x)^{1/2}}=0


    thanks in advance
    If,
    c(x) is a decreasing function and continous. Then I have a proof.
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    Junior Member miss_lolitta's Avatar
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    hi..ThePerfectHacker

    right..c(x) is a decreasing function and continous

    thanks so much
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    Quote Originally Posted by miss_lolitta View Post
    hi..ThePerfectHacker

    right..c(x) is a decreasing function and continous

    thanks so much
    Then it is simple.

    c(x)\geq 0
    Otherwise the expression is undefined.

    Now, if c(x) is decreasing function and continous then,
    \sqrt{c(x)} is decreasing.
    And we are told that,
    \int_1^{\infty} \sqrt{c(x)}dx
    Converges.

    That means, by the integral test,
    \sum_{k=1}^{\infty} \sqrt{c(k)}
    Converges.
    But that means that,
    \lim_{k\to \infty}\sqrt{c(k)}=0.

    Thus,
    \lim_{x\to \infty}x\cdot \sqrt{c(x)}=?
    We now that,
    \lim_{x\to \infty}\sqrt{c(x)}=0*)
    This is a L'Hopital rule problem.
    (But then we need to know that the function is differenciable).

    *)I am not sure about this. Perhaps this statement works only the other way.
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