It states as follows:
If $f\in C^1((a,\infty))$ is such that its limits exists, that is $ \hat{f}=\lim_{t \to \infty}f(t)$, then there exists a sequence $\lbrace t_n\rbrace$ with $t_n \longrightarrow \infty $ such that $f'(t_n)\longrightarrow 0$.
So, I find this (as a good proffesor of mine would say) obvious by drawing, but, still, it requieres a proof. I have attempted a proof, but I have never been good at real analysis and I feel insecure about at it.
I will now write my attempt and I would really apreciate if you could tell my what I have done worngly.
Lets consider $\lbrace t_n \rbrace _{n\geq 0}=\lbrace 0,\ 1,\ 2, \dots \rbrace $.
Then, we can say that $$\hat{f}=\lim_{t \to \infty}f(t)=\lim_{n \to \infty}f(t_n)=f(\lim_{n \to \infty}t_n)$$ as $f$ is continuous.
Because $f\in C^1$, it is derivable with continuous derivative $f'(t)$, therefore $$ \lim_{n \to \infty}f'(t_n)=f'(\lim_{n \to \infty} t_n)$$.
Then, $$ \lim_{n \to \infty}f'(t_n)=\lim_{n \to \infty} \lim_{h\to 0}\frac{f(t_n+h)-f(t_n)}{h}= \lim_{h\to 0}\lim_{n \to \infty}\frac{f(t_n+h)-f(t_n)}{h}=\lim_{h\to 0}\frac{\lim_{n \to \infty}f(t_n+h)-\lim_{n \to \infty}f(t_n)}{h}=\lim_{h\to 0}\frac{\hat{f}-\hat{f}}{h}=\lim_{h \to 0}\frac{0}{h}=0 $$ and so we conclude.
Thank you very much for your time!
If there is no such sequence, we can find an $\epsilon>0$ and an $M>0$ such that $x>M$ implies $|f'(x)|>\epsilon$. By continuity, this implies $f'$ does not change sign for $x>M$. Assume then that $f$ is positive, so in fact $f'(x)>\epsilon$ for $x>M$. (The negative case can be handled by a near-identical argument). By the mean value theorem, implies that $f(x)>f(M) + \epsilon(x-M)$ for $x>M$, so in fact $f(x) \to + \infty$.
Note that the continuity of $f'$ is not strictly needed here, since derivatives always satisfy the conclusion of the intermediate value theorem (Darboux's Theorem). We also don't really need to use the fact $f(x) \xrightarrow{x \to +\infty} C$, $C \in \mathbb{R}$. It's enough that $f$ is bounded.
Hence, we have the following stronger version of your theorem: