# Limit superior and limit inferior of a sequence

I have trouble understanding the concepts limit superior and inferior using the traditional definition. I have to solve a large problem and I'm not sure of how to do it. I will thank you for any hints or references of books or something where I can find any help:

Exercise 2.6.4. Given a sequence $(x_n)_{n=1}^\infty$ defined in $\mathbb{R}$, number $x^* \in \mathbb{R}$ is said to be its limit superior if $$\forall \varepsilon > 0, \exists n^* \in \mathbb{N} : \forall n > n^* , x_n < x^* + \varepsilon$$ and $$\forall \varepsilon > 0, \forall n \in \mathbb{N}, \exists n_0 > n : x_{n_0} > x^* − \varepsilon.$$ Number $x_* \in \mathbb{R}$ is the sequence’s limit inferior if $$\forall \varepsilon > 0, ∃n_* \in \mathbb{N} : \forall n > n_*, x_n > x_* − \varepsilon$$ and $$\forall \varepsilon > 0, \forall n \in \mathbb{N}, \exists n_0 > n : x_{n_0} < x_* + \varepsilon.$$ When they exist, these numbers are denoted, respectively, as $\operatorname{limsup}_{n\to\infty} x_n = x^*$ and $\operatorname{liminf}_{n\to\infty} x_n = x_*$.

1. Does the existence of the limit superior of a sequence guarantee that its limit inferior also exists?
2. Argue that if $\operatorname{limsup}_{n\to \infty} x_n = x^*$, then there exists a subsequence $(x_{n_m})$ of $(x_n)$ that converges to $x^*$.
3. Argue that, when they both exist, $\operatorname{limsup}_{n\to\infty} x_n \ge \operatorname{liminf}_{n\to\infty} x_n$.
4. Give an example of a sequence for which the previous inequality is strong.
5. Argue that if $\lim_{n\to\infty} x_n = x$, then $\operatorname{limsup}_{n\to\infty} x_n = x$.
6. Argue that if $$\operatorname{limsup}_{n\to\infty} x_n = x = \operatorname{liminf}_{n\to\infty} x_n,$$ then $\lim_{n\to \infty} x_n = x$.
• Jul 1, 2018 at 17:12
• Limit superior and limit inferior always exists for infinite sequences. It can be $+\infty$ or $-\infty$. Jul 1, 2018 at 17:20
• My opinion: It's better to define $\limsup x_n$ as the largest subsequential limit of $(x_n).$ I think that's much more intuitive. Same idea for $\liminf x_n$.
– zhw.
Jul 1, 2018 at 17:24
• @ramanujan $\pm \infty$ don't really fit into the given definition. Jul 1, 2018 at 17:44
• After you look at Lord Shark's link, I suggest you press "edit" to see how I've formatted your question. Formatting is important; it was nigh unreadable before. Jul 1, 2018 at 17:47

There are already questions out there which give some equivalent definitions and explanations about $\limsup$ and $\liminf$, for example here. 