# Prove that if a sequence converges then $\lim{x_n} = \lim \sup {x_n}$ or $\lim{x_n} = \lim \inf {x_n}$

Given a convergent sequence $$\{x_n\}$$ prove that either: $$\lim_{n \to\infty}\{x_n\} = \lim_{n \to \infty} \sup \{x_n\}$$ or $$\lim_{n \to\infty}\{x_n\} = \lim_{n \to \infty} \inf \{x_n\}$$

I believe this problem has been solved several times here, but i couldn't find such a question (probably due to translation issues, since the original problem is in another other).

I've started with gathering what is given in the problem statement. So we have that a sequence is convergent, thus: $$\lim_{n\to\infty}x_n = L \iff \{ \forall\varepsilon >0, \exists N\in \mathbb N:\forall n> N \implies |x_n-L|<\varepsilon \}$$

Also we have that the sequence is bounded, so: $$m = \inf\{x_n\} \le x_n\le \sup\{x_n\} = M \\ m \le x_n \le M$$

Now using these facts I believe I should make some assumption (for example that $$x_n$$ doesn't reach any bound and proceed by contradiction), but i can't wrap my mind for several hours already.

I would appreciate if someone could show me how to prove this or point to an already answered question.

If $$m = M$$, then the sequence is constant, so the result holds. If not, then either $$m$$ or $$M$$ (maybe both, but it doesn't matter: pick either in that case) is not equal to $$L$$. Whichever it is (call that one $$k$$), there is some $$N$$ such that for all $$n > N$$, $$|x_n - L| < \frac{|L-k|}{2}$$. Since $$k$$ is an exact bound for $$(x_n)$$, there must, for any $$\delta > 0$$ be some $$n$$ such that $$|k - x_n| < \delta$$. But for any $$\delta < \frac{|L-k|}{2}$$, this can't happen after the $$N$$th term, so must be in the first $$N$$ somewhere, so $$k$$ is an exact bound for the set of the first $$N$$ terms of $$(x_n)$$. But there are finitely many such, and every finite set achieves its exact bounds, so in particular, there is some $$n < N$$ such that $$x_n = k$$.