Ratio of two Sequences converging to zero If $\{x_{n}\}$ is a sequence of positive real numbers, $0<x_{n}<1$ and $\lim_{n\to\infty}x_{n}=0$. We Know that any subsequence of $x_{n}$ will converges to zero, right! Now my question is: Can we find (construct) a subsequence $x'_{n}$ of $x_{n}$ such that $$\lim_{n\to\infty}\frac{x'_{n}}{x_{n}}=x$$ for nonzero $x$.
(For example, if $x_{n}=\frac{1}{n}$, then we can choose $x'_{n}:=x_{2n}=\frac{1}{2n}$ and we get $\lim_{n\to\infty}\frac{x'_{n}}{x_{n}}=1/2$).
Edit: Above I said "for nonzero $x$", and I didn't specified a value for $x$, all I want is just a nonzero limit.
 A: I suppose the trivial answer to your question is "yes." After all, one can always take $n' = n$, and then $\lim_{n \rightarrow \infty} x_{n'}/x_n = 1$, since obviously each term is one. 
One possible way to make the problem less trivial is to require that $n' > n$ for all $n$. A counterexample to something like this can be given by $x_n = 1/2^{2^n}$. Note that $x_{n+1}/x_n = 2^{2^n - 2^{n+1}} = 1/2^{2^n}$, any such ratio must tend to 0.
A: Interesting question, which could get difficult if we put further conditions on $(x_n)$.  We could cheat and use a monotonically decreasing sequence $(x_n)$, and $x=3$. But let us instead use an $x \lt 1$.
Let our sequence $(x_n)$ be given by $x_n=\frac{1}{n}$ when $n$ is not a power of $2$, and $x_n=\frac{1}{2^n}$ when $n$ is a power of $2$. So $x_n$ decreases rather slowly most of the time, but occasionally takes a dramatic dip.
Let $x=\frac{1}{3}$, and suppose that we have a subsequence $(x_n')$  such that 
$$\lim_{n\to\infty}\frac{x_n'}{x_n}=\frac{1}{3}.$$
Let $m=2^k$ be a large power of $2$. 
If $m$ is sufficiently large, $\frac{x_{n+1}'}{x_{n+1}}\approx\frac{1}{3}$, so, informally, $x_n'\approx \frac{1}{3(n+1)}$.  But then we cannot have $\frac{x_n'}{x_n}$ anywhere near $\frac{1}{3}$. 
A: One thing seems to be sure: if your sequence converges monotonically to zero (from above, since we're given $\,x_n>0\,$) then any subsequence will bound it elementwise from below: $\,\,x'_n\leq x_n\,,\,\forall n\,$, and then any possible limit of the quotient of both will have to be in $\,[0,1]\,$ , so if we have $\,1<x\in\mathbb{R}\,$ we'll have to begin with a seq. that converges to zero non-monotonically, and this already rules out lots of pretty simple and basic examples, and also shows us that either we put some conditions on the sequence $\,\{x_n\}\,$ or else the answer to your question is : no, not any real $\,x\,$ can be gotten as a limit of that quotient for any sequence.
