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 $\operatorname{liminf}_{n\to\infty} x_n = x_*$.
- Does the existence of the limit superior of a sequence guarantee that its limit inferior also exists?
- 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 $x^*$.
- Argue that, when they both exist, $\operatorname{limsup}_{n\to\infty} x_n \ge \operatorname{liminf}_{n\to\infty} x_n$.
- Give an example of a sequence for which the previous inequality is strong.
- Argue that if $\lim_{n\to\infty} x_n = x$, then $\operatorname{limsup}_{n\to\infty} x_n = x$.
- Argue that if $$\operatorname{limsup}_{n\to\infty} x_n = x = \operatorname{liminf}_{n\to\infty} x_n,$$ then $\lim_{n\to \infty} x_n = x$.