$f$ twice differentiable on $(0,\infty)$, $f''$ is bounded on $(0,\infty)$, $\lim_{x\to\infty}f(x)=0$. Show that $\lim_{x\to\infty}f'(x)=0$ $f$ twice differentiable on $(0,\infty)$, $f''$ is bounded on $(0,\infty)$, $\lim_{x\to\infty}f(x)=0$. Show that $\lim_{x\to\infty}f'(x)=0$. First of all, it seems to me that the condition that $f''$ bounded is useless... Isn't it?
My attempt is to pass by mean value theorem, so i would like to know if my approach holds, please.
First, consider the interval $[x,x+a]$ with $x>0, a>0$. As $f$ is differentiable on $[x,x+a]$, by mean value theoreme there exists some $c_x \in ]x,x+a[$ such that
$f'(c_x)=\frac{f(x)-f(x+a)}{a}$.
Inserting the limit on RHS and LHS we get the following:
$\lim_{x\to\infty}f'(c_x)=\lim_{x\to\infty}\frac{f(x)-f(x+a)}{a} = 0$.
Therefore, $\lim_{x\to\infty}f'(x)=0$ as wanted.
I know there is a post already on this statement, but they use the fact that $f''$ is bounded in theirs proofs... If this hypothesis is necessary, could someone explain why or give a counter example on my proof?
 A: Counterexample. Let $f(x) = \frac 1x \sin(x^2)$. Then $f(x) \to 0$ as $x \to \infty$, but
$$
f'(x) = - \frac{1}{x^2} \sin(x^2) + \frac 1x \cos(x^2) \cdot 2x 
= - \frac{1}{x^2} \sin(x^2) + 2 \cos(x^2),
$$
which is not convergent to $0$ as $x \to \infty$. The problem here lies of course in the second derivative, which is not bounded.

Problem with your reasoning. For each $x > 0$, you find just one point $c_x \in [x,x+a]$ for which $f'(c_x)$ is small. And indeed, if you fix any sequence $x_n \to \infty$, your construction yields a corresponding sequence $c_n \in [x_n,x_n+a]$ that satisfies $c_n \to \infty$ and $f'(c_n) \to 0$. But there is no reason why every sequence $y_n \to \infty$ should be obtained as $c_n$ in this way.
In a way, this may be surprising. Your family of $c_x$ is indexed by all points $x > 0$, and the intuition may tell us that those $c_x$ should cover at least the whole halfline $[a,\infty)$. But this intuition is wrong! It would be true if $x \mapsto c_x$ was a continuous function, while it is not. The best way to see this is by studying some counterexample.

How to solve this. As you noted yourself, this is the easy part. This question was already asked and answered, e.g. here.
A: hint
Apply second order Taylor-Lagrange formula in $ [0,x] $ for an $ x >0 $.
there exists $ c\in (0,x) $ such that
$$f(0)=f(x)-xf'(x)+\frac{x^2}{2}f''(c)=0$$
