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Math Help - Another limit proof

  1. #1
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    Lightbulb Another limit proof

    Prove if lim(x sub n)=x and x>0 then there exists a natural number M such that x sub n > 0 for all n> or equal to M

    Any hints on how to start this or what the end result should be would be greatly appreciated, thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by hayter221 View Post
    Prove if lim(x sub n)=x and x>0 then there exists a natural number M such that x sub n > 0 for all n> or equal to M

    Any hints on how to start this or what the end result should be would be greatly appreciated, thanks.
    if {\rm{lim}}_{n \to \infty}{x_n}=x,\ \ x>0 then for all \varepsilon>0 there exists an N_{\varepsilon} such that:

    |x_n-x|<\varepsilon for all n\ge N_{\varepsilon}

    Now choose \varepsilon=x/2 and you should find the result follows with M=N_{\varepsilon}.

    CB
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  3. #3
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    I'm not sure I follow, what do I do with the epsilon...sorry I have just begun to learn this and I haven't fully grasped the idea of the limit definition.
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  4. #4
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    Quote Originally Posted by hayter221 View Post
    I'm not sure I follow, what do I do with the epsilon...sorry I have just begun to learn this and I haven't fully grasped the idea of the limit definition.
    You can choose anything you like for \varepsilon, that is the definition of convergence for a sequence.

    What we do is choose a particular value to ensure that x_n is positive from N_{\varepsilon} onwards. We do this by choosing any value of \varepsilon \le x/2 then as |x_n-x|<\varepsilon for all n \ge N_{\varepsilon} we have:

    |x_n-x|<x/2

    so:

    -x/2<x_n-x<x/2

    rearranging:

    x/2<x_n<3x/2

    for all n \ge N_{\varepsilon}

    But x>0, so:

    x_n>0 for all n \ge N_{\varepsilon}

    CB
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  5. #5
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    thanks, much appreciated!
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