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.

Notice that $n_k+1 = n_{k+1}$ is false most of the times so your counterexample does not work. Take $n_k = 2^k$, then
$$ \frac{a_{n_{k+1}}}{a_{n_k}} = \frac{e^{-2^{k+1}}}{e^{-2^k}} = \exp(-2^k) \to 0
$$
