this definition is from introduction to real analysis by Robert G.Bartle forth edition section 3.4 page 83 definition 3.4.10 and theorem 3. 4.11 the author include proof of these theorem but I did understand it so I tried to prove it but he didn't even define a decreasing sequence
Definition :The limit superior of $x_n$ is the infimum of the set $V$ of $v \in \mathbb{R}$ such that $v < x_n$ for at most a finite number of $n \in \mathbb{N}$.
with this detention I want to prove that
If $\epsilon > 0$:
$1)$ there are at most a finite number of $n \in \mathbb{N}$ such that $x+\epsilon<x_n$,
$2)$ but an infinite number of $n \in \mathbb{N}$ such that $x-\epsilon<x_n$
my proof is
Case $1)$
if $v<x_n$ for at most finite amount of time then there is a number $m \in \mathbb{N}$ st $\forall n>m$ $v≥x_n$ then as $x=\inf{V}$ then $x≤v<x+\epsilon$ $\forall n>m$
and because $v≥x_n$ $\forall n>m$ then $x+\epsilon >x_n$ $\forall n>m$
then at most there is $m$ number of $x+\epsilon <x_n$
Case $2)$
as there is infinite number of $n>m$ st $v≥x_n$ then
$v-\epsilon<x_n≤v$ and we know that $x≤v$ then $x-\epsilon≤v-x$ for infinite many $n$
then it follow that $x-\epsilon<x_n$
I am pretty sure that the first case proof is right but the second is not because it implies that for all n >m this assertion always hold which I don't think is true , but I had no other idea how to prove it
The first proof is correct but, in my opinion, you can come up with something simpler if you observe that the set $ V $ has the property: if $ a \notin V $ then for all $ b < a $ we have $ b \notin V $.
For the second proof I don't think your claim $ v−\epsilon <x_n \le v $ holds for $ v \in V $. The proof can be made really simple by observing that if the second claim does not hold, then $ x - \epsilon \in V $. This is an obvious contradiction with $x$ being the infimum of $ V $.