How to use least upper bound principle to prove that monotonic functions at infinity are either infinity or real?

Question

In a comment under What is the set of functions such that any quotients of two of them at infinity is real or infinity? I was told that if for some real $N$ real function $f$ is monotonic for all $x\geq N$, then $\lim_{x\to\infty}f(x)\in\mathbb{R}\cup\{\pm\infty\}$, and this is true by the least upper bound principle. A Wikipedia search also shows that this is true but no proof was given, so I'm wondering why this is true.

Answer 1

Consider the set $A:=\{f(x):x\ge N\}$. Suppose the function is increasing on this set. Either this set is bounded from above or not.

Case 1. If it is bounded, then $A$ has an upperbound $\alpha$. Then by the completeness axiom of $\mathbb{R}$, there exists $\beta:=\sup A$ the least upperbound of the set. By a standard theorem in real analysis, $f(x)$ converges to this supremum, $\beta$. (Consider any $\epsilon>0$ and show $\beta-\epsilon<f(x)\le\beta$ for $x>x_0$.)

Case 2. If it is not bounded, then for any $\alpha$, there exists $x_0$ such that $f(x_0)>\alpha$. But then $$x\ge x_0\implies f(x)\ge f(x_0)>\alpha,$$ since $f$ is increasing. This is what is meant by $f(x)\to+\infty$ as $x\to+\infty$.

So either $f$ converges to some $\beta\in\mathbb{R}$ or to $+\infty$.

The case $f$ decreasing is similar, replacing $+\infty$ with $-\infty$ and considering the infimum instead of the supremum.