Intuition for $\lim\sup$ and $\lim\inf$ After reading several alternative definitions of $\lim\sup$ and $\lim\inf$, such as $\lim\sup$ being the supremum of the set of all subsequential limits, I'm still having trouble building the intuition for $\lim\sup$.
One thing that I feel is true, but not sure, is that $\lim\sup$ represents the greatest real number that infinitely many $a_n$ gets close to, and $\lim\inf$ represents the smallest value that infinitely many $a_n$ gets close to. Are these correct statements? If so, how would one go about showing it? 
Thanks
 A: One thing that helped me: You could try to prove that if $s$ is the $\limsup$ then for any number $s' < s$, the sequence exceeds $s'$ infinitely often. And for any number $s'' > s$, the sequence exceeds $s''$ only finitely often.
A: One intuition I certainly recommend is that the $\limsup$ is the infimum of the numbers that are passed by only finitely many $a_n$'s. More precisely,
$$\limsup a_n=\inf\{\alpha\in\mathbb R:\mbox{the set $\{a_n: a_n>\alpha\}$ is finite}\}
$$
Pick $\alpha$ satisfying this. Then, there exists $n\in\mathbb N$ such that for every $k\geq n$, $a_n\leq \alpha$. Therefore,
$$
\sup_{k\geq n} a_n \leq \alpha \Rightarrow \limsup a_n = \inf_{n\in\mathbb N}\left(\sup_{k\geq n} a_n\right) \leq \alpha,
$$
so
$$
\limsup a_n \leq \inf\{\alpha\in\mathbb R:\mbox{the set $\{a_n: a_n>\alpha\}$ is finite}\}.
$$
To prove the remaining inequality, note that if $$\beta<\inf\{\alpha\in\mathbb R:\mbox{the set $\{a_n: a_n>\alpha\}$ is finite}\},$$ then certainly $$\beta\notin \{\alpha\in\mathbb R:\mbox{the set $\{a_n: a_n>\alpha\}$ is finite}\},$$
so there are infinite many $a_n$'s greater than $\beta$, so for any $n\in\mathbb N$,
$$
\sup_{k\geq n} a_n > \beta,
$$
then taking the infimum on the $n$'s, we get
$$
\limsup a_n = \inf_{n\in\mathbb N}\left(\sup_{k\geq n} a_n\right) \geq \beta.
$$
Since this happens for any $\beta<\inf\{\alpha\in\mathbb R:\mbox{the set $\{a_n: a_n>\alpha\}$ is finite}\}$, we get that
$$
\limsup a_n = \inf_{n\in\mathbb N}\left(\sup_{k\geq n} a_n\right) \geq \inf\{\alpha\in\mathbb R:\mbox{the set $\{a_n: a_n>\alpha\}$ is finite}\}.
$$
A: Yes it can be proved that if we consider the set $S\subseteq \mathbb{\bar R}$ of all the limits of the subsequences of $a_n$ we have that
$$\max\{S\}=\limsup a_n \in \mathbb{\bar R}$$
$$\min\{S\}=\liminf a_n\in \mathbb{\bar R}$$
extending the notation/definition also to the infinity cases.
That property with bounding evauation is used to prove what $\limsup$ and $\liminf$ are.
Refer also to the related


*

*What is $\lim_{x\to\infty} \sin x$?
