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Thread: 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
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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 $\displaystyle {\rm{lim}}_{n \to \infty}{x_n}=x,\ \ x>0$ then for all $\displaystyle \varepsilon>0$ there exists an $\displaystyle N_{\varepsilon}$ such that:

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

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

    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 $\displaystyle \varepsilon$, that is the definition of convergence for a sequence.

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

    $\displaystyle |x_n-x|<x/2$

    so:

    $\displaystyle -x/2<x_n-x<x/2$

    rearranging:

    $\displaystyle x/2<x_n<3x/2$

    for all $\displaystyle n \ge N_{\varepsilon}$

    But $\displaystyle x>0$, so:

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

    CB
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  5. #5
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    thanks, much appreciated!
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