I am very bad at maths , by definition we have
$$\limsup_{n\to\infty} x_n:=\inf_n(\sup_{k\geq n}x_k)$$
where $(x_n)$ is a sequence.
I was doing this in a finite sequence to understand this better: for example with
$$(2,3,5,4)$$
we start with
$n=1$ then we have $\sup\{ 2,3,5,4\}=5$.
$n=2$ then we have $\sup\{ 3,5,4\}=5$.
$n=3$ then we have $\sup\{ 5,4\}=5$.
$n=4 $ then we have $\sup\{ 4\}=4$.
and finally $\inf\{5,5,5,4\}=4$ so in conclusion that limit is $4$ in this cases?
On
A nice way to understand limit superiors (and limit inferiors) is to use the concept of subsequences. If you take a sequence $\langle x_1,x_2,\ldots,x_n,\ldots\rangle$ you may pick a subsequence, that is, another sequence of the form $\langle x_{n_1},\ldots,x_{n_k},\ldots\rangle$ where $n_1<n_2<n_3<\cdots$ and $n_k\geqslant k$ for each $k=1,2,\ldots$. We can consider, given a sequence $S$, the set $L(S)$ that consists of all elements $x\in\Bbb R$ that are the limit of some subsequence of $S$, where we allow $+\infty$ and $-\infty$ as limits. Then $L(S)$ is always nonempty: if $S$ is unbounded, either $+\infty$ or $-\infty$ (or both) lie in $S$, if $S$ is bounded, the theorem of Bolzano-Weiertrass says some subsquence of $S$ must converge to some point, which will thus lie in $L(S)$. The limit superior of $S$ is simply $\sup L(S)$, and the limit inferior $\inf L(S)$. Using this definition, one can show in fact that these are also limits of subsequences of $S$, thus by definition, $\limsup S$ is the largest number obtained as a limit of a subsequence of $S$ and $\liminf S$ is the least number obtained as a limit of a subsequence of $S$.
If your example was an infinite sequence that was eventually constant
$6, 3, 7, 9, 2, 4, 4, 4, \ldots$
then the limit would be $4$. Existence of limits with $n \to \infty$ involves 'eventual' behaviour. A sequence with only a finite number of points in cannot have limit points. In $\{2, 3, 5, 4\}$, if you get really close to $4$, meaning that you can have an open set in $\mathbb R$ for example the interval $(3.9, 4.1)$ which is small enough that it only contains $4$, it does not contain an infinite number of points of your sequence still close to $4$. So it's an isolated point. And every point in a finite set of numbers will be isolated.
For an infinite example, where we can apply your statement, consider $\{1/1, 1/2, \ldots, ..., 1/n, \ldots\}$. No matter how small an interval you place around zero, e.g. $(-1/m, 1/m)$ for any $m$ you choose, there will always be an infinite number of points from the sequence inside that interval.
Without this property, what you're trying to apply makes no sense. If
$$\limsup_{n\to\infty}x_n=\inf_n(\sup_{k\geq n}x_k)$$
then, if we choose a value of $n$, say $n=m$, $x_m = 1/m$ and $\forall k\geq m$ we have $1/m \geq 1/k$. So the $\sup_{k\geq m}$ at this point is $1/m$ whereas the infimum over all $n$ of $1/n$ is zero because $1/n\geq0, \forall n$ and there is no number greater than zero for which this is true.
So zero is the smallest of all possible values, the $\inf_n$, of the largest of the values remaining, the $\sup_{k\geq n}$ as $n$ approaches infinity, because the largest of those values, $1/n$, is approaching zero.
I hope this explanation helps. This stuff will seem quite technical at first and does take some time to get used to it.
Note: Just to point out how difficult this can be to explain, zero is not strictly the smallest value, as it is not attained by any of the remaining values in the sequence. It is just the largest value which is always less than those values remaining, now matter how large $n$ becomes. I used that wording to emphasise why $\limsup$ is equivalent to the $\inf$ of the $\sup$, by comparing the idea of smaller with larger.
Looking at another example $\{y_n=1-1/n\}$. Choose an $n$ say $n=m$. Now we have $\sup_{k\geq m}y_k=1$ as $1\geq 1-1/k, \forall k\geq m$ and no number smaller than $1$ will do. Here the $\inf_n(\sup_{k\geq n}y_k)$ is $1$ for all $n$.
Similar interpretations apply to the 'opposite' definition
$$\liminf_{n\to\infty}x_n=\sup_n(\inf_{k\geq n}x_k).$$
A good way to see the difference between $\limsup$ and $\liminf$ is to try them out on the sequence defined by
$$\left\{x_n=\begin{cases}1/n & n \text{ odd}\\1-1/n & n \text{ even}\end{cases} \ \left| \ \right.n\geq 1\right\}$$
Here, the $\liminf$ is $0$ and the $\limsup$ is $1$.