Does $f'(x) > 0$ and $f''(x) < 0$ imply that $f'(x)$ is converging to $0$ as $x$ is tending to infinity? I have a question related to another question which I asked today, where some people gave me two nice counterexamples.
Suppose you have a differentiable function $f:\mathbb{R}_+ \rightarrow \mathbb{R}_+ $, where $f'(x)>0, f''(x) <0$ and $f(0) = 0$. Does this imply that $f'(x)$ is converging to 0 as $x$ is tending to infinity?
 A: Here's a counterexample. 
Start with the function 
$$f(x) = \sqrt{x^2-1}
$$ 
on the domain $x \ge 1$; this is the upper half of one branch of a hyperbola which is asymptotic to $y=x$.
Next, translate this downward by a unit:
$$g(x) = \sqrt{x^2-1}-1
$$
so now it intersects the $x$-axis at the point $x=\sqrt{2}$.
Finally, translate this leftward by $\sqrt{2}$:
$$h(x) = \sqrt{(x + \sqrt{2})^2 - 1} - 1
$$
This function is asymptotic to a certain line of slope $+1$ (by taking the line $y=x$, translating doward by a unit, and leftward by $\sqrt{2}$). So, its derivative approaches $1$ as $x \to \infty$. And it satisfies $f(0)=0$, $f'(x)>0$, $f''(x) < 0$.
A: A reasonably straightforward counterexample is
$$f(x) = x - e^{-x} + 1$$
We have
\begin{align}
f(0) &= 0 \\
f'(x) &= 1 + e^{-x} > 0 \\
f''(x) &= -e^{-x} < 0
\end{align}
but
$$f'(x) \to 1 \neq 0 \text{ as } x \to \infty$$
A: How about $f:(1,\infty) \to (0,\infty)$ defined by $f(x) = x + \log(x)$? This has
$$ f'(x) = 1 + \frac{1}{x} > 0$$
and
$$f''(x) = -\frac{1}{x^2} < 0 $$
but 
$$ \lim_{x\to\infty} f'(x) = 1$$
