There is a theorem saying that:
$\limsup x_n = \sup{z_k}$ where ${z_k}$ is a set of limit points for the sequence ${x_n}$.
All sets and sequences are real.
The definition of limit superior is: $$\limsup x_n = \lim_{n \to \infty} \sup_{m \ge n}\{x_m\}=\inf_{n} \sup_{m \ge n}\{x_m\}$$
Limit point for a sequence is a point such that in any neighborhood of it there is a point from the sequence other than the original point itself.
If $\lim \sup x_n$ is finite, then clear enough it is a limit point itself, and there might be no other limit point which exceed $\lim \sup x_n$.
But how to proceed with the proof if $\lim \sup x_n= \pm \infty$?
Suppose $\limsup x_n=\infty$, by the $\inf \sup$ definition it means $\sup_{m\ge n} x_m=\infty$ for all $n$, in particular for $n=1$ it gives $\sup x_n=\infty$.
And, if $\limsup x_n=-\infty$, we need to prove that $x_n$ has no other limit point, i.e. $x_n\to -\infty$.
For this, take any $A\in\Bbb R$ and as $\lim_n(\sup_{m\ge n}x_m )= -\infty$, there is a limit index $N_0$ such that $n\ge N_0\implies \sup_{m\ge n}x_m<A$, and it follows that for all $m\ge N_0$, we have $x_m<A$. $\ $ -QED-
We proved a slightly stronger theorem: $$\limsup x_n=\max \{\text{limit points}\}\,.$$