Say I have a sequence {${x_n}$} that converges to $x$, that I know is consisting of nonzero, real numbers. Then can I say, by Q being dense in R, that there exists a rational number q, such that for all $x_n$ , $0<q<|x_n|$?
I need this for a larger proof.
Thanks
You are are assuming each $x_n\neq0$. If you also assume $x\neq0$, then such a $q$ does exist.
Proof: Let $\epsilon=x/2$. Note that $x_n\to x$ implies that $|x_n|\to |x|$ (proof this!).
Therefore, there is an $N$ so that for all $n\ge N$, $||x|-|x_n||<\epsilon$, which implies $|x|/2\le |x_n|$ when $n\ge N$. Now, let $y=\min(|x_1|,|x_2|,\dots,|x_N|,|x|/2)$. Then it follows that $y\le |x_n|$ for all $n$, not just $n\ge N$. Therefore, choosing any rational number in the interval $(0,y)$ would suffice.