It is clear from this question that the derivative of a $C^1$ monotone function $f : \mathbb{R} \rightarrow \mathbb{R}$ whose limit as $x \rightarrow \infty$ exists need not have the existence of a limit. I conjecture though that $f'$ must be bounded, though, but I cannot prove it, so I would be very grateful for some help, or a counterexample.
My (not particularly useful) attempt so far (this may be disregarded of course):
Without loss of generality, suppose $f$ is increasing so that $f' \ge 0$. Suppose $f'$ is not bounded. Then there exists a sequence of numbers $x_n$ such that $$\lim_{n \rightarrow \infty} x_n = \infty \\ \lim_{n \rightarrow \infty} f'(x_n) = \infty$$ (the first limit is due to the fact that $f'$ is continuous and thus bounded on compact sets). Without loss of generality, pass to a subsequence, also denoted $x_n$, such that $x_{n+1} - x_n \ge 1$. Since $f(x)$ has a limit, say $L$, as $x \rightarrow \infty$, $$\lim_{n \rightarrow \infty} f(x_n) = L$$ By the mean value theorem, there exists $y_n \in (x_n, x_{n+1})$ such that $$0 \leq f(x_{n+1}) - f(x_n) = f(y_n) (x_{n+1} - x_n) \xrightarrow{n \rightarrow \infty} 0$$ so that $f'(y_n) \xrightarrow{n \rightarrow \infty} 0$ and I thought about trying to relate $f'(y_n)$ and $f'(x_n)$ but this idea is trash, as is my math ability lol.
