Show that every subsequence converging to the same limit implies limit of sequence exists. I need to prove that if $\exists$ subsequences $a_{n_{k_{1}}}$ and $a_{n_{k_{2}}}$ (of $a_{n}$) that converge to different limits, then the sequence $a_{n}$ does not converge. I'm not sure how to do this. If I suppose $a_{n_{k_{1}}}$ converges to some limit $l_{1}$, then $\forall \epsilon >0$, $\exists n_{k_{1}}>N$ s.t. $|a_{n_{k_{2}}}-l_{1}|<\epsilon$, and the same thing for $a_{n_{k_{2}}}$ and its limit $l_{2}$. 
$a_{n}$ not converging means $\exists \epsilon_{0}$ s.t. $\forall N\in \mathbb{N}$, $\exists n_{k}\geq N$ s.t. $|a_{n}-l|\geq \epsilon_{0}$. 
But, I'm not sure how to put them together in order to prove the implication that I want.

Suppose $a_{n_{k_1}}\to a$ and $a_{n_{k_2}}\to b$, and $a<b$. Then $$\liminf_{n\to\infty}a_n \leqslant a < b \leqslant \limsup_{n\to\infty}a_n, $$
so $\lim_{n\to\infty} a_n$ does not exist. 
