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Prove that if $x > 0$ and $x_n$ is a sequence with $\lim\limits_{n \to \infty} x_n = x$, then there is a real number $N$ s.t. whenever $n > N$, $x_n > 0$.

This is a homework question and I'm not really sure what methods I should use to prove this, can I get a push in the right direction? I am not expecting a flat out answer as it is a homework problem, but I am stumped!

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Are you sure you have got the question right? As stated the claim is false: for example, the sequence $x_n=0$ is a counterexample. – Chris Eagle Oct 31 '11 at 15:43
Probably the first premise should have been "if $x>0$" rather than "if $x=0$". – Henning Makholm Oct 31 '11 at 15:48
Yes sorry, it should have been "if $x > 0$". I fixed the OP. – Matt Nashra Oct 31 '11 at 15:52
@MattNashra: also fixed title. I take out formulas if you don't mind - seems to be more readable. – Ilya Oct 31 '11 at 15:56
@MattNashra: Well in terms of $\epsilon$s and $\delta$s, what does $\displaystyle\lim_{n \to \infty} x_n = x$, mean exactly? – JavaMan Oct 31 '11 at 16:21
up vote 8 down vote accepted

It's not necessary true: consider $x_n = -1/n$. Though it would be true if $x>0$.

Hint: consider the interval $(x-\epsilon,x+\epsilon)$ for $\epsilon = x/2$. Recall the definition of $\lim\limits_{n\to\infty}x_n=x$ in $N$-$\epsilon$ terms.

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