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Math Help - Prove that there is a threshold k such that xn>0 for all n>=k.

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    Prove that there is a threshold k such that xn>0 for all n>=k.

    Let (x_n) be a convergent sequence with limit a. Assume a>0. Prove that there is a threshold k such that x_n>0 for all n \geqslant k.

    What I have:

    Let \varepsilon >0. Then there is a k such that for all n \geqslant k, |x_n-a|< \varepsilon . Assume a>0. Then since \exists k such that ( \forall n \geqslant k, |x_n-a|< \varepsilon . We know that 0<|[x_n-a|< \varepsilon which is a<|xn|< \varepsilon +a. Since a>0 we have |xn|>a>0.
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    Quote Originally Posted by alice8675309 View Post
    Let (x_n) be a convergent sequence with limit a. Assume a>0. Prove that there is a threshold k such that x_n>0 for all n \geqslant k.
    Pick \varepsilon  = \frac{a}{2}.
    If |x_n-a|<\frac{a}{2} what do get when rewritten without absolute values signs?
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    Quote Originally Posted by Plato View Post
    Pick \varepsilon  = \frac{a}{2}.
    If |x_n-a|<\frac{a}{2} what do get when rewritten without absolute values signs?
    \frac{a}{2}<xn< \frac{3a}{2} ? Which means that if \frac{a}{2}= \varepsilon then xn> \frac{a}{2}= \varepsilon ? so if xn> \varepsilon and \varepsilon >0 so xn>0?
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    That does it. Right?
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    Quote Originally Posted by Plato View Post
    That does it. Right?
    yup. My question now is, when you look at a something to prove how do you know when to choose epsilon to be something other than just greater than 0 like you did in the problem by letting epsilon be a/2?
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    There is really no way to answer that question, other than to say one word: experience.
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