# 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

• Well both lim sup and lim inf can be infinite, so perhaps you should say "extended real number." Nov 7, 2018 at 5:10
• yes, that's what I meant. Thanks for pointing out.
– gws
Nov 7, 2018 at 5:14
• Which is the definition of $\limsup$ that you are using? Nov 7, 2018 at 5:14
• Yes, $\limsup$ is the largest possible limiting value achievable over a subsequence of times that go to infinity, while $\liminf$ is the smallest possible. Intuition comes by considering $\limsup_{t\rightarrow\infty} \cos(t) = 1$ and $\liminf_{t\rightarrow\infty} \cos(t) = -1$. We have $\limsup = \liminf = \lim$ if and only if the regular limit $\lim$ exists. Nov 7, 2018 at 5:27
• You may have a look at this answer math.stackexchange.com/a/1893725/72031 Nov 8, 2018 at 13:31

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.

• I tried and I think I proved this. But I was just wondering would s be the unique real number that satisfies these two properties?
– gws
Nov 7, 2018 at 6:38
• Yes, it is. Any $t< s$ fails to satisfy the property: "for any number $u>t$, the sequence exceeds $u$ only finitely often" (to show that, just pick $u\in(t,s)$, since $u<s$, the sequence must exceed $u$ infinitely often). Analogously, we prove that any $s<t$ fails to satisfy the property: "for any number $u<t$, the sequence exceeds $u$ infinitely often". So $s$ is the only one that satisfies both those properties. Nov 7, 2018 at 10:25

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 $$\limsup a_n = \inf_{n\in\mathbb N}\left(\sup_{k\geq n} a_n\right) \geq \inf\{\alpha\in\mathbb R:\mbox{the set \{a_n: a_n>\alpha\} is finite}\}.$$

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