Question: Let $\{x\_n\}\_{n\in \mathbb{N}}\subset \mathbb{R}$ be a bounded sequence. We define $\forall N \in \mathbb{N}$:
$$a\_N := inf\{x\_n \mid n \geq N\}$$
$$b\_N := sup\{x\_n \mid n \geq N\}$$
$$l := \lim\_{N \to \infty} a\_N$$
$$L := \lim\_{N \to \infty} b\_N$$
1. Prove that $\{x\_n\}\_{n\in \mathbb{N}}$ converges $\iff$ $l = L$.  

2. Prove that there exist subsequences such that one converges to $l$ and the other to $L$.

I've proved $a\_N$ is monotonic increasing, $b\_N$ is monotonic decreasing, that $a\_N \leq x\_N \leq b\_N$, $\forall N\in \mathbb{N}$, and that $l\leq L$. It seems that the right-to-left impliation of question 1 can be proved using the sandwich rule...

I'd like a full answer to question 1 and hint(s) for question 2.

Answer: 
If $(x\_n)$ converges, say to $p$, then the idea is clearly that $l = L = p$ in this case.
Suppose $l < L$ instead. Then take $\epsilon < \frac{L - l}{2}$, then all terms of the sequence $x\_n$ where $n >= K$, for some $K$, lie in $(p-\epsilon, p+\epsilon)$. What does this say about $a\_K$ and $b\_K$?

Consider $l-1$ first. The $a\_n$ increase to $l$, so from some index $m\_1$, $a\_{m\_1} > l-1$, so some $x\_{n\_1}$, with $n\_1 \ge m\_1$, $x\_{n\_1} > l-1$. Now find $m\_2 > n\_1$ with $a\_{m\_2} > l-\frac{1}{2}$ and $n\_2 \ge m\_2$ with $x\_{n\_2} > l - \frac{1}{2}$, and so on, using $l-\frac{1}{k}$. The sequence of $x\_{n\_1}, x\_{n\_2}, \ldots$ converges to $l$. Similarly for $L$, of course.
2) consider the sequence $a\_N+1/N$ resp. $b\_N-1/N$ and their relations to the partial sequence $x\_N,x\_{n+1},...$.

The second part of 1) is actually a consequence of 2), since all subsequences of a convergent sequence have the same limit.