Limit Supremum and Infimum. Struggling the concept

Let's first just recall the definitions of the limit superior and limit inferior. For a sequence $\{a_n\}$, they are $$\limsup_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} \sup_{k\geq n} a_k, \quad \liminf_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} \inf_{k \geq n} a_k.$$ Recall that the supremum is the least upper bound and the infimum is the greatest lower bound. So then the expressions $\sup_{k \geq n} a_k$ and $\inf_{k \geq n} a_k$ are the upper and lower bounds for the tails of the sequence, looking only at terms $k \geq n$.

So the limit superior is asking, how large can the tails of the sequence eventually be? Similarly, the limit inferior is asking, how small can the tails of the sequence eventually be?

Example: Let $a_n = \{100, -100, -1, 1, -1, 1, -1, 1, \ldots\}$. Then $\sup_n a_n = 100$, $\inf_n a_n = -100$, but the $\limsup$ ignores all the large terms that begin in the finite portion of the sequence, so we have $\limsup a_n = 1$. Similarly, the $\liminf$ ignores all the small terms in the beginning, so $\liminf a_n = -1$.

To address the example you gave in your question, the sequence you gave is not an infinite sequence so $\limsup$ and $\liminf$ aren't defined.

I'd like to add to Christopher A. Wong's answer that the $\liminf$ is the smallest accumulation point while the $\limsup$ is the largest one. Moreover, if you have an understanding of the $\liminf$ already, then consider $\limsup a_n = -\liminf (-a_n)$.