Let $f:(a, \infty)\to \Bbb R$ be a differentiable function such that exists $\lim_{x\to\infty}f(x)=l<\infty$ and exists (in the sense it can also be infinity) $\lim_{x\to\infty}f'(x)$. Under these hypothesis we can conclude that $\lim_{x\to\infty}f'(x)=0$.
I found a proof which says the following: by Lagrange theorem for every $x$ there exists a number $\eta_x$ in between $x$ and $x+1$ such that $f(x+1)-f(x)=f'(\eta_x)$. So we conclude that $\lim_{x\to\infty}f'(\eta_x)=\lim_{x\to\infty}(f(x+1)-f(x))=l-l=0$.
Is this really a proof? I suspect it is wrong as a proof. Should we ensure that there exists such a continuous function $x\mapsto\eta_x$? In fact, for example, $\eta$ could assume only integer values, because between $x$ and $x+1$ there always is an integer $n$. So, in this case we could only conclude that $\lim_{n\to\infty}f'(n)=0$ (where $n$ is an integer), and not that $\lim_{x\to\infty}f'(x)=0$ (where $x$ is any real number).
How can we "save" this proof?
As pointed out by David Mitra, you don't need continuity of the map $x\mapsto \eta_x$, but only the fact that $\eta_n\to +\infty$ as $n$ goes to infinity (because $\eta_n\geqslant n$). Since it is assumed that $\lim_{x\to +\infty}f'(x)$ exists, and is equal to (say) $L$, we have $\lim_{n\to +\infty}f'(\eta_n)=L$, and we conclude that $L=0$ by your computations.