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Math Help - Showing Upper Limt= Lower Limit

  1. #1
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    Showing Upper Limt= Lower Limit

    Hey, can anyone help in this theorem.

    If {Xn } is abounded sequence,then its limit exist as (n goes to infinity) if and only iff

    Lim supXn = Lim infXn
    n-->infinity n-->infinity


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  2. #2
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    i'm thinking we may use squeeze theorem here. but again, not getting it.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by concern View Post
    Hey, can anyone help in this theorem.

    If {Xn } is abounded sequence,then its limit exist as (n goes to infinity) if and only iff

    Lim supXn = Lim infXn
    n-->infinity n-->infinity


    Use that \liminf_{n\to\infty}x_n\le\lim_{n\to\infty}x_n\le\  limsup_{n\to\infty}x_n to prove the one part. Next consider instead of limsup,\liminf think about it as \sup\text{ }X,\inf\text{ }X where X is the set of all subsequential limits of \left\{x_n\right\}. And use the fact that if every subsequential limit converges to the same value that the limit exists and equals that value.
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    you are great Drexel28. Your help is truly awesome. In my 1st post also you were the only one who could come up with genuine solution. Hats off to you.

    Quote Originally Posted by Drexel28 View Post
    Use that \liminf_{n\to\infty}x_n\le\lim_{n\to\infty}x_n\le\  limsup_{n\to\infty}x_n to prove the one part. Next consider instead of limsup,\liminf think about it as \sup\text{ }X,\inf\text{ }X where X is the set of all subsequential limits of \left\{x_n\right\}. And use the fact that if every subsequential limit converges to the same value that the limit exists and equals that value.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by concern View Post
    you are great Drexel28. Your help is truly awesome. In my 1st post also you were the only one who could come up with genuine solution. Hats off to you.
    There are a lot of great members on here. I just think that I have conicidentally answered a large amount of yours.
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