Real Analysis - differentiable $f:[0,\infty]\rightarrow \mathbb{R}$ is twice differentiable. If $f''$ is bounded and exists the limit of $f(x)$ at infinity, then $\lim_{x\rightarrow \infty}f'(x)=0$.
I tried to use the Taylor's formula but I couldn't prove that $\lim_{x\rightarrow \infty}f'(x)=0$.
 A: I assume that the limit of $f$ is finite...
$\forall x>0$, there exists $c_x\in(0,x)$ such that
$$f(0)=f(x)+f'(x)(0-x)+\frac{f''(c_x)}{2}(0-x)^2$$
Use this equation to write $f'(x)=\dots$ and use your hypothesis about $f$ and $f''$.
A: If $\displaystyle \lim_{x \to \infty}f(x) = L$ is finite and $\displaystyle \lim_{x \to \infty}f'(x) = L'$ exists, then it is always the case that $L' = 0$.
Note that by the MVT for some $\xi_x \in [x,x+1]$,
$$f(x+1) = f(x) + f'(\xi_x)$$
and
$$L' = \lim_{x \to \infty}f'(x) = \lim_{x \to \infty}f'(\xi_x)=\lim_{x \to \infty}[f(x+1)-f(x)]=0.$$
However, there are cases where the limit of $f(x)$ is finite, but the limit of $f'(x)$ fails to exist. As an example, consider $f(x) = 1 + \sin(x^2)/x$.
Some further work is required to determine if the boundedness of the second derivative precludes this possibility.
A: Hint: If you are saying that $\lim_{x \to \infty} f(x)$ exists and is finite, then assume $|f'(x)| \geq \epsilon$ for some fixed $\epsilon > 0$ for infinitely many arbitrarily large $x$, and that $|f''(x)| < C$ always. Then choose an infinite increasing sequence of such $x_n$ so that $|f'(x)| \geq \epsilon$, and $x_{n+1} - x_n > 1$. Then over a small interval starting at $x_n$, say of width $\delta$, the change in $f$ from $x$ to $x + \delta$ will essentially be $\delta*f'(x_n)$. If $f''$ is bounded and $|f'(x_n)| \geq \epsilon$ then you can come up with a lower bound for $|\delta*f'(x)|$ over the interval if $\delta$ is small enough, and thus come up with a lower bound for the change in $|f|$ over the interval. This is all you need.
