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

Question

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.

Answer 1

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 $x_n = k$.