Find a subsequence whose limit exists 
Let $\{a_n\}$ be a sequence of non-zero real numbers.
Show that it has a subsequence $\{a_{n_k}\}$ such that $\lim \dfrac{a_{n_{k+1}}}{ {a_{n_k}}}$ exists and belongs to $\{0,1,\infty\}$.

I am finding the above problem false.
If I take $(a_n)_n=(e^{-n})_n$ then any sub-sequence of $a_n$ is $e^{-n_k}$ but $\lim \dfrac{a_{n_{k+1}}}{ {a_{n_k}}}=\dfrac{e^{-n-1}}{e^{-n}}=\dfrac{1}{e}\notin \{0,1,\infty\}$.
Edits:By @Henry's comment I am sure the problem is true.But how should I  find the sub-sequence.Please give some hints.

If the sequence $(a_n)_{n\geq0}$ is unbounded then WLOG we may assume that for each $N\geq0$ there is an $n$ with $a_n>N$. In this case let $n_0$ be the first $n$ with $a_n>0$, and choose for any $k\geq1$  an $n_k>n_{k-1}$ with $a_{n_k}\geq k \>a_{n_{k-1}}$. It follows that $\lim_{k\to\infty}{a_{n_k}\over a_{n_{k-1}}}=\infty$.
If the sequence $(a_n)_{n\geq0}$ is bounded by Bolzano's theorem there is a subsequence $(a_{n_k})_{k\geq0}$ converging to some $\lambda\in{\mathbb R}$. If $\lambda\ne0$ then $\lim_{k\to\infty}{a_{n_k}\over a_{n_{k-1}}}=1$. If $\lambda=0$ then after sieving we may assume $\lim_{n\to\infty} a_n=0$, and a second sieving allows to assume that, e.g., $a_n>0$ for all $n\geq0$. In this case  put $n_0=0$, and choose for any $k\geq1$ an $n_k$ with $0<a_{n_k}\leq{a_{n_{k-1}}\over k}$. It follows that $\lim_{k\to\infty}{a_{n_k}\over a_{n_{k-1}}}=0$.
