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Math Help - limes proof

  1. #1
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    limes proof

    hello,

    I am given a being the limes of a sequence an and s being a real number

    (a) let all elements of the seqence be >= 0, the it follows that a >= 0 as well.
    (b) if an >= s for all n, then it follows that a >= s.
    (c) if an <= s for all n, then it follows that a <= s.

    How can I arrive at (b) and (c) coming from a? I first thought about epsilon criteria, but that didn't get me anywhere. Greatly appreciating your help...

    oz.

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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by ozfingwe View Post
    hello,

    I am given a being the limes of a sequence an and s being a real number

    (a) let all elements of the seqence be >= 0, the it follows that a >= 0 as well.
    (b) if an >= s for all n, then it follows that a >= s.
    (c) if an <= s for all n, then it follows that a <= s.

    How can I arrive at (b) and (c) coming from a? I first thought about epsilon criteria, but that didn't get me anywhere. Greatly appreciating your help...

    oz.

    Oops! Misread. So given (a) consider \tilde{a}_n=a_n-s and b_n=s-a_n
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  3. #3
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    Quote Originally Posted by ozfingwe View Post
    I am given a being the limits of a sequence an and s being a real number
    (b) if an >= s for all n, then it follows that a >= s.
    Here is another way to do part (b).
    Suppose that a<s, then s-a>0.
    So \left( {\exists N} \right)\left[ {\left| {a_N  - a} \right| < s - a} \right] \Rightarrow \quad a_N  < s.
    That is a contradiction.
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