Let's say I want to compute :
$$\lim_{n \to \infty} O\left(\frac{n-1}{n}\right)$$
Then can we say that to infty this is equal to $O(1)$ ? Because it could also mean that $\forall n \in \mathbb{N}$ there is a constant $C_n$ such that : $\mid f\mid \leq C_n \cdot \frac{n-1}{n}$
And in the case where the sequence : $C_n$ is such that : $n = o(C_n)$ then we have : $$\lim_{n \to \infty} O\left(\frac{n-1}{n}\right) = O(+\infty)$$
That's why I am wondering : when we say that a function $f$ is $O(\frac{n-1}{n})$ then does it mean : $\mid f\mid \leq C \cdot \frac{n-1}{n} \forall n$, or : $\forall n, \exists C_n$ such that $\mid f \mid \leq C_n \cdot \frac{n-1}{n}$ ?
Neither. It means $\exists n_0 \exists C$ such that $\forall n \ge n_0: |f| \le C \cdot \frac{n-1}{n}$.
$O$-notation is often abused, but I've never before seen $\lim_{n \to \infty} O(f(n))$. If that makes any sense at all, it's to disambiguate between the asymptotic behaviour for large $n$ and the asymptotic behaviour for small $n$.
Finally, observe that if $n > 0$ and $|f| \le C \cdot \frac{n-1}{n}$ then certainly $|f| \le C \cdot 1$, and we can indeed say that $f(n) \in O(1)$.