The second sentence doesn’t make sense as written; I expect that you’re actually supposed to show that
if for every $\epsilon>0$ there is an $m\in\Bbb N$ such that $|a_k-a|<\epsilon$ for all $k\ge m$, then $\limsup_k a_k=a$ and $\liminf_k a_k=a$.
It may be helpful, at least in getting a feel for what’s going on here, to realize that the hypothesis
for every $\epsilon>0$ there is an $m\in\Bbb N$ such that $|a_k-a|<\epsilon$ for all $k\ge m$
says precisely that the sequence $\langle a_k:k\in\Bbb N\rangle$ converges to $a$. Thus, you’re being asked to show that if a sequence has a limit, that limit is also its limit superior and its limit inferior.
The first sentence of your last paragraph is missing several crucial bits and consequently doesn’t make sense as written. It should read:
If $a=\limsup_k a_k$, then for each $\epsilon>0$ there is an $m\in\Bbb N$ such that $a_k<a+\epsilon$ whenever $k\ge m$.
Unfortunately, this isn’t much help as it stands. You need something that implies that $\limsup_k a_k=a$, and this does just the opposite: it tells you something that is implied by the statement that $\limsup_k a_k=a$. Something can be done along these lines, but at this point I think that you’re probably better served by going back to the definition.
To state it more carefully, you know that $$\limsup_k a_k=\lim_{n\to\infty}\sup_{k\ge n}a_k\;.$$ To show that $\limsup_k a_k=a$, therefore, you have to show that for each $\epsilon>0$ there is an $m\in\Bbb N$ such that $$\left|a-\sup_{k\ge n}a_k\right|<\epsilon$$ whenever $n\ge m$.
So let $\epsilon>0$ be any positive number. What do we know that might be useful? We’re told that there is an $m\in\Bbb N$ such that $|a-a_k|<\epsilon$ whenever $k\ge m$. That’s at least the right kind of statement; does it actually help with what we’re trying to prove? Suppose that $n\ge m$; what can we say about $$\sup_{k\ge n}a_k\;?$$ If $k\ge n$, then $k\ge m$, so $|a-a_k|<\epsilon$, and therefore $a-\epsilon<a_k<a+\epsilon$. In other words, the entire tail $\{a_k:k\ge n\}$ of the sequence lies between $a-\epsilon$ and $a+\epsilon$. This implies that $$a-\epsilon<\sup_{k\ge n}a_k\le a+\epsilon$$ for every $n\ge m$. (You should try to figure out why I could write a strict inequality on the left but only a non-strict inequality on the right.) In other words,
$$\left|a-\sup_{k\ge n}a_k\right|\le\epsilon$$
for every $n\ge m$. This isn’t quite what I wanted, because I have $\le$ instead of $<$, but it’s good enough to ensure that $$\lim_{n\to\infty}\sup_{k\ge n}a_k=a$$ and hence that $\limsup_k a_k=a$. (You should take a moment to see why this is true.)
The proof for the $\liminf$ is very similar.
Yes. In fact, the limit superior of a sequence $(x_n)$ is precisely the supremum of the set $S$ of subsequencial limits of $(x_n)$, and analogously, the limit inferior is the infimum of $S$.
Also, there will always be a subsequence of $(x_n)$ converging to its limit superior, and likewise there will always be a subsequence of $(x_n)$ converging to its limit inferior. That is to say, $\limsup x_n, \, \liminf x_n \in S$.
In your example where $x_n = \cos \left(\frac{n\pi}{2} \right)$, you found a subsequence converging to $1$, and this sequence $x_n$ is never any bigger than $1$ by the nature of cosine, so $1$ is automatically the largest a subsequencial limit could be.
Best Answer
Notice that the positive terms are decreasing and the negative terms are increasing.
Hence $\sup_{k \ge n} x_k$ will be given by the first positive term in $x_n,x_{n+1},...$, and similarly, $\inf_{k \ge n} x_k$ will be given by the first negative term in $x_n,x_{n+1},...$.
Since the first positive term of the whole sequence is $2$, we have $\sup_n x_n = 2$, and the $\inf$ is computed similarly.
We have $\lim_{n \to \infty} \sup_{k \ge n} x_k = \lim_{n \to \infty} \sup_{k \ge n} x_{2k-1} = \lim_{n \to \infty} \sup_{k \ge n} (1+{1 \over k}) = \lim_{n \to \infty} (1+{1 \over n}) = 1$.
The $\liminf$ is computed in a similar fashion.
Note: In general, the sequence $n \mapsto \sup_{k \ge n} x_k$ is non-increasing, the $\limsup$ is the limit of this non-increasing sequence. Similarly for $\liminf$.