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My textbook gives the following definition of limit superior, which I'd like some clarification on.

Let $\{s_n\}$ be a sequence that's bounded above and does not diverge to $-\infty$. Then there is a unique real number $\bar{s}$ (the limit superior) such that if $\epsilon > 0$, $$s_n < \bar{s} + \epsilon$$ for "large $n$" and $$s_n > \bar{s} - \epsilon$$ for infinitely many $n$.

First of all, is the textbook saying that for the limit superior to equal $\bar{s}$, the inequalities above must hold for all $\epsilon > 0$? Secondly, what exactly does "large $n$" mean in this context?

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Your textbook is not using a very precise language. It should be: $\limsup_ns_n$ is the only real number $\overline s$ such that:

  1. for every $\varepsilon>0$, there is a $p\in\mathbb N$ such that$$(\forall n\in\mathbb{N}):n\geqslant p\implies s_n<\overline s+\varepsilon;$$
  2. for every $\varepsilon>0$, the inequality $s_n>\overline s-\varepsilon$ holds for infinitely many $n$'s.
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  • $\begingroup$ Loosely speaking, could 1. be interpreted as saying that for all $\epsilon > 0$, there exists a $p$ such that once the sequence reaches the $p$th term, the inequality is true for a given $\epsilon$? $\endgroup$ – user484604 Nov 20 '17 at 18:43
  • $\begingroup$ @Taliant Yes, if you delete the words “for a given $\varepsilon$”. $\endgroup$ – José Carlos Santos Nov 20 '17 at 18:53
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"For large $n$", or "eventually", mean "for all $n$ starting at a certain point".

And yes, when the text says "if $\epsilon>0$, then this, then mean that you can do it for every $\epsilon>0$ (the starting point where the inequality holds will change with $\epsilon$, though).

I have always felt that a much better intuition for $\limsup$ is to take it as the supremum of all limits of all subsequences.

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