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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 $\lim_{n\to \infty} x_n = x$.
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    $\begingroup$ Please see math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Angina Seng Jul 1 '18 at 17:12
  • $\begingroup$ Limit superior and limit inferior always exists for infinite sequences. It can be $+\infty$ or $-\infty$. $\endgroup$ – ramanujan Jul 1 '18 at 17:20
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    $\begingroup$ 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 $\liminf x_n$. $\endgroup$ – zhw. Jul 1 '18 at 17:24
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    $\begingroup$ @ramanujan $\pm \infty$ don't really fit into the given definition. $\endgroup$ – Theo Bendit Jul 1 '18 at 17:44
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    $\begingroup$ 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. $\endgroup$ – Theo Bendit Jul 1 '18 at 17:47
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There are already questions out there which give some equivalent definitions and explanations about $\limsup$ and $\liminf$, for example here.

Personally, I like to think that the sequence eventually becomes "wedged" between the limsup and liminf, as illustrated in the picture below. The picture should give you some good intuition for your questions 2-6.

enter image description here

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  • $\begingroup$ To the proposer: There are many Q's with answers on this site, on this topic. $\endgroup$ – DanielWainfleet Jul 1 '18 at 17:53

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