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Prove that if lim (Xn) = x and if x>0, then there exists m from N Xn>0 for all n>=m.

  1. Mar 10, 2012 #1
    1. The problem statement, all variables and given/known data
    I'll try to restate it more clearer (I'm sorry I'm not good with using the more mathematical text type.

    Prove that if lim (Xn) = x and if x>0, then there exists a natural number M such that Xn > 0 for all n≥M.


    2. Relevant equations
    Just to be clear, this limit definition is assumed to be to infinity as per our book for analysis.
    (Xn) represents the sequence notation
    And Xn will represent my term for a given n.

    Scratch work pieces are that I want to choose an ε = x (the limit) at term XM
    This would imply that for all n ≥ M, 0<Xn<2x

    I would like to approach this using an ε-neighborhood.


    3. The attempt at a solution
    Assume that the lim(Xn) = x and x > 0.
    => Given any ε>0, |Xn-x| < ε
    => -ε + x < Xn < ε + x
    => There exists a natural number M s.t. 0 < XM < 2x (This is from the scratch work)
    => For all n ≥ M, 0<Xn<2x
    => Xn > 0 for all n ≥ M


    I do want to make sure this is right. I realize that it is sloppy currently and I would love to have more help in writing this a bit nicer. It makes sense to me, but it's weird to choose a term that "fits" that epsilon.

    Thanks for your help!
     
  2. jcsd
  3. Mar 10, 2012 #2

    LCKurtz

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    Re: Prove that if lim (Xn) = x and if x>0, then there exists m from N Xn>0 for all n>

    OK, since your argument is in fact mathematically correct, I will show you how to better present it. My comments will be interspersed in red.

     
  4. Mar 11, 2012 #3
    Re: Prove that if lim (Xn) = x and if x>0, then there exists m from N Xn>0 for all n>

    Thank you very much! Sorry for the late response. I just want you to know that the help is greatly appreciated, and the flow makes it much easier to read!
     
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