Do the lim sup and lim inf always exist? Let's focus on the $\lim \inf$. While trying to prove that the $\lim \inf$ always exists, [this page]https://mathcs.org/analysis/reals/numseq/proofs/lub_ex.html established that the sequence $A_j = \inf\{a_j, a_{j+1}, \dots\}$ is monotone increasing then the $\lim \inf$ exists even though it may be possibly infinite. 
My question is that if we define the limit of a sequence as a real number as [the same page]https://mathcs.org/analysis/reals/numseq/sequence.html has done then how do we reconcile this definition to allow for "a limit that is infinite"?
 A: The presentation is just a little sloppy, in that the authors have not previously explained that a sequence that increases (decreases, resp.) without bound is said to converge to $+\infty$ ($-\infty$, resp.), but the idea should be clear anyway. It is not really contradictory: it is just an extension of the earlier definition.
And it is entirely compatible with that definition once you know a bit of topology. One forms the extended real numbers by adjoining points $-\infty$ and $+\infty$ and giving them nbhds analogous to $\epsilon$-nbhds $(a-\epsilon,a+\epsilon)$ around real numbers. Specifically, just as we say that a sequence $\langle x_n:n\in\Bbb N\rangle$ converges to $a$ if and only if for each $\epsilon>0$ there is an $n_0\in\Bbb N$ such that $x_n\in(a-\epsilon,a+\epsilon)$ for all $n\ge n_0$. we say that it converges to $+\infty$ if and only if for each $b\in\Bbb R$ there is an $n_0\in\Bbb N$ such that $x_n>b$ for all $n\ge n_0$. The open rays 
$$(b,\to)=\{+\infty\}\cup\{x\in\Bbb R:x>b\}$$
function as the equivalent at $+\infty$ of open intervals $(a-\epsilon,a+\epsilon)$ at $a$. And just as at $a$ it generally suffices to work just with the countable family of intervals $\left(a-\frac1n,a+\frac1n\right)$, so at $+\infty$ we can generally work just with the countable family of rays $(m,\to)$ for $m\in\Bbb Z$.
The notion of convergence of a sequence to $-\infty$ is defined similarly: $\langle x_n:n\in\Bbb N\rangle$ converges to $-\infty$ if and only if for each $b\in\Bbb R$ there is an $n_0\in\Bbb N$ such that $x_n<b$ for all $n\ge n_0$.
From a topological point of view convergence to $\infty$ or $-\infty$ behaves no differently from convergence to some $a\in\Bbb R$; both are instances of a more general notion of sequential convergence. And it is convenient to make the extension even at this early point, because sequences that diverge in $\Bbb R$ because they converge to $\infty$ or $-\infty$ in the extended reals, rather than for other reasons (like $\langle (-1)^n:n\in\Bbb N\rangle$, for instance) are quite well-behaved and can often be treated right along with those that converge to real numbers.
