How do I rigorously show a sequence of positive real numbers converges to a non-negative real number? 
If $x_n >0$ is a sequence of positive real numbers and $x_n\to x$ as $n \to \infty$, prove that $x \ge 0$.

Suppose $x<0$. Since $x_n \to x$, then for all $\epsilon$ there exists $N$ such that $n \ge N \implies |x_n-x|< \epsilon$.
I know we must have infinitely many terms of the sequence close to $x$, but does this say there must exist infinitely many negative terms in the sequence? If so, how?
 A: Assume that $x<0$, then there exists some $N$ such that if $n\geq N$, then $|x_{n}-x|<-x/2$, then $x_{n}=x_{n}-x+x\leq|x_{n}-x|+x<-x/2+x=x/2<0$.
A: If $x$ is negative and $x_n$ is given to be positive (or more generally non-negative) then $|x_n-x|\geq |x|$ and there is no way to make the difference $|x_n-x|$ smaller than $|x|$ and this contradicts the fact that $x_n\to x$. Therefore we must have $x\geq 0$.
A: The precise form of "...infinitely many terms close to $x\;$" is that $\lim_{n\to \infty}x_n=x$ iff for every $e>0$ the set $S(e)=\{n: x_n\not \in (-e+x,e+x)\}$ is finite. 
If $\lim_{n\to \infty}x_n=x<0$ then with $e=|x|/2,$ the set $S(|x|/2)=\{n: x_n\not \in (-3|x|/2,-|x|/2)\}$ is finite, which implies that $\{n: x_n> 0\}$ is a subset of the $finite$ set $S(|x|/2),$ so it cannot be true that $x_n>0$ for all $n.$
A: There is another way of looking at this using Topology.
Since $\mathbb R$ is a metrizable space, then it is first-countable and thus it is sequential.
The set $S = [0, + \infty) \subset \mathbb R$ is a closed set in $\mathbb R$ and so any convergent sequence from $S$ will have its limit lying inside $S$.
Now since your convergent sequence by assumption lies inside $\{ x \in \mathbb R \ | \ x \gt 0\} \subset S$ then its limit must be contained in $S$ by the above logic.
This implies that the limit is non-negative.
A: Because every negative number has a neighborhood consisting of negative numbers only.
