The relevant definitions can be found here:
$\liminf_{s\to\infty}\Gamma(s)=\bigcup_{s\ge 1}\left(\bigcap_{k\ge s}\Gamma(k)\right)\;,\tag{1}$
$\limsup_{s\to\infty}\Gamma(s)=\bigcap_{s\ge 1}\left(\bigcup_{k\ge s}\Gamma(k)\right)\;,\tag{2}$
and $\displaystyle\lim_{s\to\infty}\Gamma(s)$ exists iff $\displaystyle\liminf_{s\to\infty}\Gamma(s)=\limsup_{s\to\infty}\Gamma(s)$, in which case $\displaystyle\lim_{s\to\infty}\Gamma(s)=\liminf_{s\to\infty}\Gamma(s)$.
It follows from $(1)$ that $x\in\liminf\limits_{s\to\infty}\Gamma(s)$ iff $x$ is in all of the sets $\Gamma(s)$ from some $s$ on:
$x\in\liminf_{s\to\infty}\Gamma(s)\quad\text{iff}\quad\exists m\in\Bbb Z^+\text{ such that }x\in\Gamma(s)\text{ for all }s\ge m\;.$
It follows from $(2)$ that $x\in\limsup\limits_{s\to\infty}\Gamma(s)$ iff $x$ belongs to infinitely many of the sets $\Gamma(s)$:
$x\in\limsup_{s\to\infty}\Gamma(s)\quad\text{iff}\quad\forall m\in\Bbb Z^+~\exists s\ge m\text{ such that }x\in\Gamma(s)\;.$
It follows that $\lim\limits_{s\to\infty}\Gamma(s)$ exists iff each point that belongs to infinitely many of the $\Gamma(s)$ belongs to all of them from some point on; apparently this is not the case, however.