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Math Help - Advanced Calculus - Limit Proof

  1. #1
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    Advanced Calculus - Limit Proof

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

    My thoughts
    Intuitively, this makes sense. If as n -> \infty, x_n=x>0 then this means that at some point n \geq M where x_n>0.

    So far I've gotten:

    Given \epsilon > 0 then,
     |x_n-x|<\epsilon
    -\epsilon<x_n-x<\epsilon
    -\epsilon + x <x_n<\epsilon + x
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Luxury View Post
    Prove that if lim(x_n)=x and if x>0, then there exists a natural number M such that x_n>0 for all n \geq M.

    My thoughts
    Intuitively, this makes sense. If as n -> \infty, x_n=x>0 then this means that at some point n \geq M where x_n>0.

    So far I've gotten:

    Given \epsilon > 0 then,
     |x_n-x|<\epsilon
    -\epsilon<x_n-x<\epsilon
    -\epsilon + x <x_n<\epsilon + x
    Since x\in(0,\infty) which is an open set there exists some \delta>0 such that B_{\delta}(x)\subseteq(0,\infty) and since x_n\to x there exists some N\in\mathbb{N} such that N\leqslant n\implies x_n\in B_{\delta}(x). Form you conclusion.
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    Quote Originally Posted by Drexel28 View Post
    N\leqslant n\implies x_n\in B_{\delta}(x)
    Could you clarify that part? I'm not sure how you went from left to imply right.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Luxury View Post
    Could you clarify that part? I'm not sure how you went from left to imply right.
    You may not be familiar with the open ball notation. Well, how about letting \varepsilon=\frac{x}{2}>0 so there exists some N\in\mathbb{N} such that N\leqslant n\implies \left|x_n-x\right|<\frac{x}{2}\implies x+\frac{-x}{2}<x_n and so...
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  5. #5
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    Lim(x_n)=x and x>0

    Since x>0, it is a little easier to simply let x=epsilon.
    Substitute in x for epsilon in you inequality.
    And since  n \geq K(epsilon) from the definition of the  lim(x_n)=x , let K(epsilon) = M.
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