Since this old question was reopened by a new answer, I will give my input as I believe an important property is missing from all the answers provided so far.
Let $\{a_n\}\subseteq \mathbb{R}$ be a sequence of real numbers.
We often study the limit of $a_n$ when $n$ goes to infinity in order to deduce more properties about the sequence itself. This value is denoted with $\lim_{n\to\infty} a_n$. The problem with the limit is that, sometimes, it might not exist.
A simple example is the sequence
$$
a_n=(-1)^{n},
$$
which oscillates between $-1$ and $1$. It is easy to show, with the epsilon delta (or any other) definition of the limit, that this sequence does not have one.
However, it does have two subsequences that converge, the sequence of even-indexed elements which converges to $1$, and the sequence of odd-indexed elements which converges to $-1$. This is not a coincidence.
The $\limsup$ and $\liminf$, unlike the limit itself, always exist, and this is perhaps their main and most important property. There are two main things that we need to show this.
- You might know that for any set $S\subseteq \mathbb{R}$, there always exists a supremum of $S$, which is either a real number or $\pm\infty$, or similarly the infimum. In other words,
$$
\text{for every $S\subseteq \mathbb{R}$ we have $\sup(S),\inf(S) \in \mathbb{R}\cup \{\pm\infty\}$}.
$$
- The second important property to know is that if any $\{b_n\}\subseteq\mathbb{R}$ is a monotonic sequence (might be increasing or decreasing), then the limit of $b_n$ exists, and is a member of $\mathbb{R}\cup\{\pm \infty\}$. (I consider the limits going to $\infty$ or $-\infty$ as well defined "existing" limits. The sequence may not converge to a real number, but its divergence towards $\infty$ (or $-\infty$) is a well defined behaviour which gives us a lot of information.)
Now finally we get to the question of the $\limsup$ and the $\liminf$.
By the definition that you provided,
$$
\limsup_{n\to\infty} a_n = \lim_{N \to \infty} \sup \{a_k\:|\: k \geq N\}.
$$
Let's consider the sequence ${s_N}$ where $s_N=\sup \{a_k\:|\: k \geq N\}$.
\begin{align*}
s_0&=\sup\{ a_0, a_1, a_2, a_3, \ldots \}, \\
s_1&=\sup\{\phantom{ a_0, }\, a_1, a_2, a_3, \ldots \}, \\
s_2&=\sup\{\phantom{ a_0, a_1 }\,\,\, a_2, a_3, \ldots \}, \\
s_2&=\sup\{\phantom{ a_0, a_1, a_2 }\,\,\, a_3, \ldots \}, \\
\,\,\,&\,\,\,\vdots
\end{align*}
Suppose, for the sake of simplicity, that for each of these sets $A_N=\{a_k\:|\: k \geq N\}$, their supremum is contained in the set. This is not true, far from it, but let's just assume it is.
Then the supremum $s_N$ is the largest element of $A_N$.
What does this tell us about $s_{N+1}$?
Well, since $s_N$ is the largest element of $A_N$, $s_{N+1}$ he largest element of $A_{N+1}$, and $A_{N+1}$ is smaller than $A_N$, then $s_{N+1}$ must be at most as large as $s_N$, but not larger. In other words $s_{N+1} \leq s_{N}$.
Or, more generally, $s_0\geq s_1 \geq \cdots$, that is
$$
\text{$s_N$ is a }\textit{monotone decreasing} \text{ sequence. }
$$
Therefore, it has a limit. Therefore the $\limsup_{n\to\infty}a_n$ exists !
In full generality, even if $s_N\not\in A_N$, we can show that for any $S \subseteq T$ we have $\sup(S) \leq \sup(T)$ (I will leave this as an exercise). Then $s_N \geq s_{N+1}$ because $A_N \supseteq A_{N+1}$.
The argument is the same for $\liminf_{n\to\infty}a_n$, except that $S \subseteq T \implies \inf(S)\geq \inf(T)$, so the sequence of infimums is an increasing sequence, again monotonic, and thus still converging.
