# Prove that $x_n$ ∈ $]x_*-e,x^*+e[$ for all n>N and $x_*$=lim inf$(x_n)$ and $x^*$=lim sup$(x_n)$

Let $$x_n$$ be a bounded sequence of real numbers, let $$x_*$$=lim inf$$(x_n)$$ and $$x^*$$=lim sup$$(x_n)$$. Also, let e > 0. Prove that there exists an N in the naturals such that $$x_n$$$$]x_*-e,x^*+e[$$ for all n>N. I wanted to go by contradiction. So, finding a subsequence that is never in the interval. But how would I do that?

Assume there exists an $$\epsilon >0$$ for which you have for every $$N \in \mathbb{N}$$ at least one $$n>N$$ such that $$x_n > x^* + \epsilon$$.
Now recall the definition: $$\limsup\limits_{n\to\infty} x_n := \inf\limits_{N\in\mathbb{N}} \sup\limits_{n \geq N} x_k$$
What can you say about $$\sup\limits_{n \geq N} x_n$$?