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Math Help - series limit proove question..

  1. #1
    MHF Contributor
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    series limit proove question..

    there is a continues function f(x) and bounded on (x_0,+infinity)
    proove that for every T there is a sequence
    X_n=+infinity
    so
    lim [f(x_n +T) - f(x_n)]=0
    n->+infinity

    i tried to solve it like this:
    the bound is L
    if f(x) is bounded than there is e>0 |f(x)-L|<e
    but i dont know what to do next
    ??
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by transgalactic View Post
    there is a continues function f(x) and bounded on (x_0,+infinity)
    proove that for every T there is a sequence
    X_n=+infinity
    so
    lim [f(x_n +T) - f(x_n)]=0
    n->+infinity

    i tried to solve it like this:
    the bound is L
    if f(x) is bounded than there is e>0 |f(x)-L|<e
    but i dont know what to do next
    ??
    I am interpreting this that if f(x) and x_n\to \infty prove that for any T, \lim_{n\to\infty}\left[ f\left(x_n+T\right)-f(x_n)\right]=0

    Suppose that \lim_{x\to\infty}f(x)=a. So we know that given any \varepsilon>0 there exists a \eta>0 such that \eta< x\implies |f(x)-a|<\varepsilon~(1), and since x_n\to \infty we may find a \delta>0 such that \delta<n\implies \eta<x_n~(2). Now suppose that T>0 (the proof for the other cases is analgous), then choose \delta such that (2)\implies (1) then \left[f\left(x+T\right)-f(x)\right|\leqslant \left|f\left(x+T\right)-a\right|+|f(x)-a|<2\varepsilon, this implies the result.


    Does that help you at all?

    Note: If this was a wrong interpretation of your question or this is unclear feel completely free to ask
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  3. #3
    MHF Contributor
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    i learned from the delta proofes article that when you define the delta

    <br />
\delta>0<br />
    it needs to come with[/COLOR]
    <br />
|x-x_3|<\delta<br />
    in our case x_3 goes to infinity
    so the inqueality that i presented not logical

    but on the other hand
    it how its done on the article limit proove

    ??
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