Consider a sequence of functions $f_k:(0,\infty)\mapsto(0,\infty)$ and a rate sequence $r_k \downarrow 0$. Assume that the functions are uniformly bounded from above by a constant $M$, i.e. $\Vert f_k \Vert_\infty<M$, and that, for all positive constants $C>0$, $\lim_{k\to \infty}f_k(C)/r_k$ exists (finite). Can we then conclude that there exists a sequence $C_k\to \infty$ which diverges "not too fast" such that $f_k(C_k)/r_k=O(1)$?
I was trying to derive a positive answer by contradiction, but I got stuck.
Let $n(i)$ be increasing such that $r_k < \frac{1}{i}$ if $k > n(i)$.
For $k \in [n(i), n(i + 1))$, define $$f_k(x) = \begin{cases} r_k \cdot x, & x < k\\ r_k \cdot k, & x \geq k \end{cases}$$
Then each $f_k(x) \leq 1$.
$f_k(x) / r_k = x$ for $x < k$ so $\lim\limits_{k\to \infty}f_k(C) / r_k$ exists for any $C$.
For $x \geq r_k$ we have $f_k(x) / r_k = k$. Thus we have $f_k(x) / r_k \geq \min(k, x)$.
So for any $C_k \to \infty$ we have $f_k(C_k) / r_k \geq \min(k, C_k) \to \infty$ and thus $f_k(C_k) / r_k \neq O(1)$ for any $C_k \to \infty$.