Suppose $f:(0, \infty)\to\mathbf{R}$ is bounded (by $B$, say) and non-negative. Then $$l := \limsup_{x\to\infty} f(x)$$ exists. The proof I'm reading now claims that $f(x) \le l + o(1)$, which I'm taking to mean "there exists some $o(1)$ function $g$ such that $f(x) \le l + g(x)$ for all $x \ge 0$." Is this the correct interpretation? If so, what could this function $g$ be?
I know that the definition of limit supremum implies the following: for all $\epsilon > 0$ there is an $x_0$ such that $f(x) < l + \epsilon$ for all $x \ge x_0$. Can we construct an explicit $g$ by getting a bunch of $x_n$'s, going to infinity, chosen with the property that $f(x) < l + 1/n$ for all $x \ge x_n$, and then defining $g$ as piecewise constant? e.g. in terms of indicator functions, $$g(x) = B\chi_{(0, x_0)}(x) + \sum_{n\ge 0}\frac{1}{n}\chi_{[x_n, x_{n+1})}(x).$$
The notation "$ p(x)=o(q(x))$ as $x\to \infty$", when $q(x)\ne 0$ for all sufficiently large $x$, means that $\lim_{x \to \infty}|p(x)/q(x)|=0$.For any $q$ it means $$\forall k>0\exists r (|p(x)|\leq k|q(x)|).$$ In particular when $q(x)=1$ it means $\lim_{x\to \infty}|p(x)|=0$. So $f(x)\leq l+o(1) \iff f(x)-l|\leq o(1)\iff f(x)-l \leq q(x)$ for some $q$ such that $\lim_{x\to \infty}|q(x)|=0.$ Which is just another way of saying $\lim_{r\to \infty}\sup \{f(x):x>r\}\leq l$. For other examples we have $\ln x=o(x)$ and $\sqrt {n^2+2 n}-(n+1)=o(1)$as $x\to \infty$. $$\\$$Note: "$o(q(x))$" is read "little o $q(x)$" or "little o of $q(x)$" (or "small o") to distinguish it from "big O". If you see $$\\$$"$a(x)=O(b(x))$ as $x\to \infty$" it means $\exists k>0 \exists r ( \forall x>r (|a(x)|\leq k|b(x)|).$ For example $99 n=O(n)$ as $n\to \infty.$