Converse of "positive sequence converges to positive limit" It is well known, and not hard to prove, that if a convergent real sequence $(x_n)$ is eventually positive (i.e. there is some $K$ for which $n > K \Rightarrow x_n > 0$), then the limit to which it converges must be non-negative. I am interested in whether the converse is true: that is, if we know that $(x_n)$ converges to a non-negative limit, then is it true that there exists $K$ with the property stated above?
 A: Let $(x_n)$ converges to $x(> 0)$. i.e.,
For a given $\epsilon>0, \hspace{1mm}\exists K\in \Bbb{N}$ , such that $\forall n\geq K, x_{n}\in (x-\epsilon, x+\epsilon)$. This works for any $\epsilon>0$.
Let the given $\epsilon:=x>0 \implies \exists K\in \Bbb{N}$ , such that $\forall n\geq K, x_{n}\in (x-\epsilon, x+\epsilon)=(0,2x)$ OR $0<x_{n}<2x$.
Hence, there exists a $K$ for which $x_{k}>0$ for all $n\geq K$.
A: If $(x_n)$ converges to a strictly positive limit $x$, then the terms of the sequence must eventually become positive. By definition of convergence, there is a $K \in \mathbb{N}$ such that $|x_n - x| < x$ for all $n \geq K$ (using $x$, which is greater than zero, as the "$\varepsilon$"). In particular, this implies that for all $n \geq K$, we have $-(x_n - x) < x$, or equivalently, $x_n > 0$.
However, if a sequence converges to zero then it need not be eventually positive.
A: Well the big picture is if $\lim a_n \to a > 0$ then eventually the distance between $a_n$ and $a$ will get very small and it will eventually become smaller than $a$ itself and if $a_n$ is closer to $a$ then $a$ is to $0$ then $a_n$ must be positive (if $a_n \le 0$ then $a_n$ would be at least as far away as $a$ is from $0$).
Can you prove it formally with the definition?

$\lim_{n\to \infty} a_n = a$ means that for any $\epsilon > 0$ there is a $K$ so that whenever $n > K$ then $|a_n - a| < \epsilon$.

What should we choose $\epsilon$ to be to make $a_n$ closer to $a$ than $a$ is to $0$?

 If $a > 0$ what if we choose  $\epsilon \le a$?  Then there is a $K$ so that whenever $n > K$ then $|a_n - a| < \epsilon \le a$.  So $-a < a_n - a < a$ so $0 < a_n < 2a$.  So $a_n>0$ for all $n >K$.  

