Consider this example:
$$
3-\frac12,\quad 5+\frac13,\quad 3-\frac14,\quad 5+\frac15,\quad 3-\frac16,\quad 5+\frac17,\quad 3-\frac18,\quad 5+\frac19,\quad\ldots\ldots
$$
It alternates between something approaching $3$ from below and something approaching $5$ from above. The lim inf is $3$ and the lim sup is $5$.

The inf of the whole sequence is $3-\frac12$.

If you throw away the first term or the first two terms, the inf of what's left is $3-\frac14$.

If you throw away all the terms up to that one and the one after it, the inf of what's left is $3-\frac16$.

If you throw away all the terms up to that one and the one after it, the inf of what's left is $3-\frac18$.

If you throw away all the terms up to that one and the one after it, the inf of what's left is $3-\frac1{10}$.

. . . and so on. You see that these **inf**s are getting bigger.

If you look at the sequence of **inf**s, their **sup** is $3$.

Thus the lim inf is the **sup** of the sequence of **inf**s of all tail-ends of the sequence. In mathematical notation,
$$
\begin{align}
\liminf_{n\to\infty} a_n & = \sup_{n=1,2,3,\ldots} \inf_{m=n,n+1,n+2,\ldots} a_m \\[12pt]
& = \sup_{n=1,2,3,\ldots} \inf\left\{ a_n, a_{n+1}, a_{n+2}, a_{n+3},\ldots \right\} \\[12pt]
& = \sup\left\{ \inf\left\{ a_n, a_{n+1}, a_{n+2}, a_{n+3},\ldots \right\} : n=1,2,3,\ldots \right\} \\[12pt]
& = \sup\left\{ \inf\{ a_m : m\ge n\} : n=1,2,3,\ldots \right\}.
\end{align}
$$

Just as the lim inf is a **sup** of **inf**s, so the lim sup is and **inf** of **sup**s.

One can also say that $L=\liminf\limits_{n\to\infty} a_n$ precisely if for all $\varepsilon>0$, no matter how small, there exists an index $N$ so large that for all $n\ge N$, $a_n>L-\varepsilon$, and $L$ is the largest number for which this holds.