If $\lim_{x\to\infty}f(x)$ and $\lim_{x\to\infty}f^{\prime}(x)$ both exist, then $\lim_{x\to\infty}f^{\prime}(x) = 0$ Suppose $f:\mathbb{R}\to\mathbb{R}$ is everywhere differentiable, and suppose that $\lim_{x\to\infty}f(x)$ and $\lim_{x\to\infty}f^{\prime}(x)$ both exist. I am trying to prove that the latter limit is necessarily $0$. I have the following argument, but I'm not sure if it's completely sound.
Since $f$ is differentiable everywhere, we can apply the Mean Value Theorem to $f$ on $[x,x+1]$ for all relevant $x$. This guarantees an $\alpha_{x}\in(x,x+1)$ such that $$f^{\prime}(\alpha_{x}) = \frac{f(x+1)-f(x)}{x+1-x} = f(x+1)-f(x).$$ Now, the limit as $x\to\infty$ of the right-hand side of this expression must be $0$, since $\lim_{x\to\infty}f(x)$ exists by assumption (and must equal $\lim_{x\to\infty}f(x+1)$). On the left hand side, we notice that $\alpha_{x}\to\infty$ as $x\to\infty$, since $\alpha_{x}>x$ always, so that:
\begin{eqnarray*}
0 & = & \lim_{x\to\infty}[f(x+1)-f(x)]\\
& = & \lim_{x\to\infty}f^{\prime}(\alpha_{x})\\
& = & \lim_{y\to\infty}f^{\prime}(y),
\end{eqnarray*}
proving the result.
I took inspiration for this argument from other sources which use the same trick of "use the Mean Value Theorem to introduce a quantity $\alpha_{x}$ which we have some bounds on, then take limits". However, this style of argument seems dodgy to me: we haven't actually defined a function $\alpha$ to take the limit of as $x\to\infty$, and it's not clear to me that defining such a function is always possible. For example, we can't just say "take the least such value and call it $\alpha_{x}$", because we haven't shown that there will always be a least such value.
Here are my questions:


*

*In the above, where have we used the fact that $\lim_{x\to\infty}f^{\prime}(x)$ exists? This is an important assumption: consider for example the function $x\mapsto\sin{(x^{2})}/x$. My guess is that it's used in the last line, where we must assume this fact to use the chain rule, but I'd like confirmation of this.

*Does the "$\alpha_{x}$ trick" require something like the Axiom of Choice in general? In particular, the thing which makes me slightly anxious about just saying "choose an $\alpha_{x}$ for every $x$" is that we have to make (uncountably) infinitely many "choices", and we have no prescribed method of doing this. EDIT: It turns out this has been answered in other questions on this site, see link in the comments below.
EDIT: Note that the first question is different to others on related topics because here I am asking very specifically about this argument and why it works.
 A: As for your first question: the mean value theorem only guarantees the existence of some $\alpha_x$ with the given property, but you have no control where $\alpha_x$ is exactly found in the interval. The fact that $f^\prime$ converges allows you to to conclude that this does not matter, any sequence $x_n$ converging to $\infty$ will have the property $f^\prime(x_n)\rightarrow \lim_{x\rightarrow\infty} f^\prime(x)$, in particular the one you get using the MVT.
I don't get what you want to know with your second question. The proof you found is rigorous.
A: For every $n$, by the mean value theorem, you have $f(n+m)-f(n)=mf'(x_{n,m})$. Since $lim_{x\rightarrow +\infty}f(x)=l$ exists, there exists $N_n$ such, $m>N_n$ implies $\mid f(n+m)-l\mid <1$, so $\mid f(n+m)-f(n)\mid =m\mid f'(x_{n,m})\mid< \mid l|+1+\mid f(n)\mid$ we deduce that
$\mid f'(x_{n,m})\mid \leq {{1+\mid l\mid +\mid f(n)\mid}\over m}$. This implies that for every $n>0,$ for  every $c>0$ there exists $x_{n,m_0}=u_n>n$ such that $\mid f'(u_n)\mid <{1\over c}$. Remark that the limit of $u_n$ is $+\infty$ and the limit of $f'(u_n)$ is zero. Since $lim_{x\rightarrow+\infty}f'(x)$ exists, we deduce that this limit is zero.
A: The only property of $\alpha_{x}$ which matters here is that $x<\alpha_{x}<x+1$ and we don't need to know details of how $ \alpha_{x}$ depends on $x$. We know that $f(x+1)-f(x)$ tends to $0$ and hence given any $\epsilon>0$ there exists an $M >0$ such that $$|f(x+1)-f(x)|<\epsilon\tag{1}$$ whenever $x>M$. Futhere we know that $f'(x)$ tends to $L$ as $x\to \infty$ so that there is a number $N>0$ such that $$L - \epsilon <f'(x)<L +\epsilon\tag{2}$$ whenever $x>N$. It is now obvious by mean value theorem that there is an $\alpha_{x}\in(x,x+1)$ such that $$-\epsilon<f'(\alpha_{x})<\epsilon\tag{3}$$ whenever $x>M$. If $L\neq 0$ then we can't have both inequalities $(2)$ and $(3)$ simultaneously for $\epsilon = |L|/2$ and $x > \max(M,N)$. Hence $L=0$.
