Let $a, \alpha \in \Bbb{R}$; let $f: (a,+\infty)\to \mathbb{R}$ be differentiable; let $\lim_{x\to \infty}f(x)=\alpha$; let $\beta := \lim_{x\to \infty}f'(x)$. I want to show that $\beta = 0$. Now, the idea is quite clear to me.
If $\lim_{x\to \infty}f(x)=\alpha$ then there is one horizontal assymptote and thus $f$ will get closer to it as $x$ grows. In that case, since it is horizontal, the derivative of $f$ should vanish.
Although the idea is clear I'm not being able to write down this proof. I've tried by contradiction: suppose $\beta > 0$ first. Then since $\lim_{x\to \infty}f(x)=\alpha$, given $\epsilon > 0$ we have $|f(x)-\alpha|<\epsilon$ as long as $x > M$ for some suitable $M\in \mathbb{R}$, $M > 0$.
Now, if we pick $x > M$ and focus on the interval $(M,x)$ there should be $c$, by the mean value theorem, such that
$$f'(c) = \dfrac{f(x)-f(M)}{x-M}.$$
In this way I can introduce the derivative there, but it is computed at a fixed point. Of course if I vary $x$ the $c$ would vary and I believe I could get a contradiction from this, but I'm not finding how to do this properly.
So how can I write down this idea for this proof? How can I get one contradiction from this?
Consider (as proposed by David Mitra in the comments) $$ f(n+1)-f(n)=\frac{f(n+1)-f(n)}{(n+1)-n}=f'(n+\theta_n),\qquad \theta_n\in(0,1) $$ Now we know that the limits on both sides exist by the assumptions of the statement. The limit on the left is $α-α=0$, the limit on the right $β$ as $c_n=n+\theta_n\to\infty$.