# If $\{x_n\}$ is a monotone sequence and there exists $k$ such that $\lim_{n \rightarrow \infty} x_n = x_k$, then $x_n=x_k$ for $n>k$

Let $\{x_n\}$ be a convergent monotone sequence. Suppose that there exists a $k \in N$ such that $$\lim_{n \rightarrow \infty} x_n = x_k$$ Show that $x_n = x_k$ for all $n \ge k$. I am completely lost in knowing where to begin this problem.

Hint: suppose $x_m\neq x_k$ for some $m>k$. Can your sequence still be monotone?
Solution: Suppose $\displaystyle\lim_{n\rightarrow\infty}x_n=x_k$ for some $k$. Now, suppose that there exists an $m>k$ such that $x_m\neq x_k$ - say $x_m<x_k$, WLOG. Set $\epsilon=x_k-x_m$. Now, because the sequence converges to $x_k$, there must be a $j>m>k$ such that $\vert x_k-x_j\vert<\epsilon$, or written another way, $$x_k-\epsilon<x_j<x_k+\epsilon.$$ By construction, $x_m=x_k-\epsilon$, and so we have $x_j>x_m$. However this means that the sequence cannot be monotone, since for $k<m<j$ we have $x_k>x_m$ while $x_m<x_j$ (i.e. the sequence decreases, then increases again). The same proof will work if we supposed that $x_m>x_k$.
Well, the alternative is that $x_n > x_k$ for some $n > k$. What does this imply with regards to the relation between the limit and $x_k$, given that the sequence is monotone?