If $f''$ is bounded, does $\lim_{x\to\infty} f'(x)$ exists? Let $f:[0,\infty)\rightarrow \mathbb{R}$ twice differentiable and such that $f''$ is bounded, and $\lim_{x\to\infty} f(x)$ exists. Does $\lim_{x\to\infty} f'(x)$ exists?
How do i guarantee that?
 A: Yes.
Assume $|f''(x)|<M$ for all $x$ and $\lim_{x\to\infty}f(x)=a$.
Let $\epsilon>0$ be given. Then also $\tilde\epsilon=\frac{\epsilon^2}{4M}>0$ and from the given convergence, there exists $x_0>0$ such that $|f(x)-a|<\tilde\epsilon$ for all $x> x_0$.
Let $x>x_0$. Assume $f'(x)\ge\epsilon$, then $f'(x+t)>\epsilon-Mt$ for all $t>0$ and hence $f(x+t)-f(x)>\epsilon t-\frac12 Mt^2$ for all $t>0$. 
In particular, for $t=\frac\epsilon M$, we have
$$f(x+t)-f(x)> \epsilon t-\frac12Mt^2=\frac{\epsilon^2}{2M}=2\tilde\epsilon,$$
contradicting $|f(x+t)-f(x)|\le |f(x+t)-a|+|a-f(x)|<\tilde\epsilon +\tilde\epsilon$. Hence the assumption $f'(x)\ge \epsilon$ must be wrong, i.e., $f'(x)<\epsilon$ for all $x>x_0$. By the same argument, we show $f'(x)>-\epsilon$ for all $x>x_0$.
Together this shows: For all $\epsilon>0$ there exists $x_0>0$ such that $|f'(x)|<\epsilon$ for all $x>x_0$. In other words,
$$ \lim_{x\to\infty}f'(x)=0.$$
A: Here is one way to proceed: suppose $f'$ does not have a limit as $x \to \infty$. Then the Cauchy property fails: there exists $\epsilon > 0$ with the property that for every $n \in \mathbb N$ there exist $x_n,y_n \ge n$ satisfying $$|f'(x_n) - f'(y_n)| \ge \epsilon.$$ Thus for each $n$ either $|f'(x_n)| \ge \epsilon/2$ or $|f'(y_n)| \ge \epsilon/2.$ Relabel if necessary to obtain a sequence $x_n \to \infty$ satisfying $|f'(x_n)| \ge \epsilon/2$ for all $n$.
At this point the graph of $f$ has a bunch of little bumps as it approaches its limit. What does the second derivative have to contribute? It implies that the bumps aren't in fact so little. The mean value theorem and the boundedness of $f''$ give you a constant $M$ with the property that $$|f'(x_n) - f'(z)| \le M|x_n - z|$$ for each $n$. In particular, if $|x_n - z| < \epsilon/(4M)$ then $|f'(z)| > \frac{\epsilon}4.$
This is getting too long, but this means that on each subinterval of the form $$\left( x_n - \frac{\epsilon}{4M},   x_n + \frac{\epsilon}{4M} \right)$$ you have either either $f' > \epsilon/4$ or $f' < -\epsilon/4$. Thus $$ \left| f\left(x_n + \frac{\epsilon}{4M} \right) - f\left(x_n - \frac{\epsilon}{4M} \right) \right| \ge \frac{\epsilon^2}{16M}$$ for all $n$ which is inconsistent with $f$ having a limit.
