Given any bounded sequence $x(n)$, let $l$ be its limit superior. Let $\epsilon > 0$.

  1. Does there exist $n \in \mathbb{N}$ such that $x(n)<l+\epsilon$?
  2. Does there exist infinitely many $n$ such that $l-\epsilon < x(n)$?
  • $\begingroup$ what is $x(n)$? $\endgroup$
    – Thomas
    Jun 18, 2014 at 14:17
  • $\begingroup$ You're right, @Hayden. Poor choice of a word, on my part. It's just that the OP usually has no idea why his/her post failed to render, so doesn't know how to correct it. It's a bug I had to figure out over the course of a year, noting why that happened, and when, to finally realize that it happens only with "<" and then only when there is no space following it. $\endgroup$
    – amWhy
    Jun 18, 2014 at 14:29
  • $\begingroup$ @amWhy that's very true, I made the bad assumption of assuming the question was gibberish without actually looking to see if it didn't render correctly. Thanks, I'll keep in mind the bug for the future! $\endgroup$
    – Hayden
    Jun 18, 2014 at 14:33
  • 1
    $\begingroup$ It had never occurred to me to think of what the plural is, but I'm pretty sure it would never have occurred to me to think it might be "limit superiors". I'd have written "limits superior". Don't others find "limit superiors" surprising and counterintuitive? $\endgroup$ Jun 18, 2014 at 16:21

2 Answers 2


In fact, one of several equivalent definition which are commonly used for limit superior is that:

The number $l$ is called limit superior if

  • for each $\varepsilon>0$ there exists $n_0$ such that for $n\ge n_0$ the inequality $x_n\le l+\varepsilon$ holds;
  • for each $\varepsilon>0$ and for each $n_0$ there exists $n\ge n_0$ such that $x_n\ge l-\varepsilon$.

In symbols: $$ (\forall\varepsilon>0)(\exists n_0)(\forall n\ge n_0) x_n\le l+\varepsilon\\ (\forall\varepsilon>0)(\forall n_0)(\exists n\ge n_0) x_n\ge l-\varepsilon $$

So if this was definition which your book/course uses for limit superior, the properties follow directly from the definition.

If you are using another definition, you may try to prove the equivalence of two definitions of limit superior as an exercise. (BTW you might have mentioned what is your definition of $\limsup x_n$ in your post.) In the definition which I have mentioned above, the case that $\limsup x_n=+\infty$ must be dealt with separately.

See also these posts, where several commonly used definitions of limit superior are mentioned:


1) Yes. If to the contrary you have $x(n) \ge l + \epsilon$ for all $n$ then also you will have $l \ge l + \epsilon$, which can't happen.

2) Yes. If to the contrary you have $l - \epsilon < x(n)$ for at most finitely many $n$, there is a point in the sequence beyond which $l - \epsilon \ge x(n)$ for all $n$. This would force $l - \epsilon \ge l$, which can't happen.


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