Show that if $\lim\limits_{x \to \infty} f(x)$ exists and $f''$ is bounded, then $\lim\limits_{x \to \infty} f'(x)=0$. I'm trying to answer the following exercise: Suppose that $f$ is twice differentiable on $[0,+\infty)$. 

Show that if $\lim\limits_{x \to \infty} f(x)$ exists and $f''$ is bounded, then $\lim\limits_{x \to \infty} f'(x)=0$.
  If $\lim\limits_{x \to +\infty} f'(x)$ exists then I know how to prove that this limit is equal to 0 (by using the mean value theorem). 
But how can I show that this limit exists? 

Counterexample: Let us consider  $f(x) = x \sin\frac{1}{x}$. Then $$f'(x) = \sin\frac{1}{x} - \frac{1}{x} \cos\frac{1}{x}$$ and $$f''(x) = -\frac{1}{x^2}\cos\frac{1}{x} + \frac{1}{x^2}\cos\frac{1}{x} - \frac{1}{x^3} \sin\frac{1}{x}.$$ This example shows that the hypothesis that there exists $\lim\limits_{x \to \infty} f(x)$ is essential. 
 A: We are given that $f''$ is bounded, so let
$$
|f''(x)|\le M\tag1
$$
Suppose that
$$
\limsup\limits_{x\to\infty}|f'(x)|=A\tag2
$$
Since $\lim\limits_{x\to\infty}f(x)$ exists, there is an $L$ so that if $x,y\ge L$,
$$
|f(x)-f(y)|\le\frac{A^2}{8M}\tag3
$$
By $(2)$, there is an $x_0\ge L+\frac{A}{2M}$ so that $|f'(x_0)|\gt\frac A2$. Then, by $(1)$, $|f'(x)|\ge\frac A2-M|x-x_0|$. Since $f'$ has the same sign over $\left[x_0-\frac{A}{2M},x_0+\frac{A}{2M}\right]$, we have
$$
\begin{align}
\left|\,f\left(x_0+\frac{A}{2M}\right)-f\left(x_0-\frac{A}{2M}\right)\,\right|
&=\left|\,\int_{x_0-\frac{A}{2M}}^{x_0+\frac{A}{2M}}f'(x)\,\mathrm{d}x\,\right|\\
&=\int_{x_0-\frac{A}{2M}}^{x_0+\frac{A}{2M}}\left|f'(x)\right|\,\mathrm{d}x\\
&\ge\int_{x_0-\frac{A}{2M}}^{x_0+\frac{A}{2M}}\left(\frac A2-M|x-x_0|\right)\,\mathrm{d}x\\
&=\frac{A^2}{4M}\tag4
\end{align}
$$
However, the only way that $(3)$ and $(4)$ do not contradict is if $A=0$. That is,
$$
\limsup\limits_{x\to\infty}|f'(x)|=0\tag5
$$
which implies
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{x\to\infty}f'(x)=0}\tag6
$$

Finiteness of $A$
Note that since $\lim\limits_{x\to\infty}f(x)$ exists, there is an $M_2$ so that for $x\ge M_2$, $|f(x)-f(x+1)|\le1$. The Mean Value Theorem says that for some $\xi\in(x,x+1)$, $|f'(\xi)|\le1$. The Mean Value Theorem and $(1)$ say that for any $\zeta\in(x,x+1)$, $|f'(\zeta)-f'(\xi)|\le M$. Therefore, the $\limsup$ mentioned in $(2)$ is finite; in fact, it is bounded by $M+1$.
A: Since $f$ is twice differentiable,  by Taylor's expansion, for every $\varepsilon>0$, there exists $c_{x,\varepsilon}\in(x,x+\varepsilon)$ such that $$f(x+\varepsilon)=f(x)+f'(x)\varepsilon+\frac{1}{2}f''(c_{x,\varepsilon})\varepsilon^2.$$
But $f''$ is bounded (say, by $ |f''|<2M$), so it follows that
$$|f(x+\varepsilon)-f(x)-f'(x)\varepsilon|\le M\varepsilon^2.$$
Therefore, by triangular inequality,
 $$|f'(x)\varepsilon|\le|f(x+\varepsilon)-f(x)|+ M\varepsilon^2.$$
But,  $\lim_{x\to\infty}f(x)$ exists, so fixing $\varepsilon>0$ and letting $x\to\infty$, we we get 
$$\limsup_{x\to\infty}|f'(x)|\le M\varepsilon.$$
Since $\varepsilon>0$ is arbitrary, the conclusion follows.
