Q:

Concave function with oblique asymptote

I'm looking for a continuous concave function $f:[0,\infty)\to [0,\infty)$ such that $f(0)=0$ and there exists some $\alpha>0$ with $\lim_{x\to \infty}\frac{f(x)}{x} =\alpha$ and $\forall x>0, f(x)>\alpha x$.

I haven't been able to come up with such a function off the top of my head. Intuitively, $f$ must have an inflection point.
Context : a microeconomics professor suggested such a function exists in a lecture

A:

For example, combine  $e^{-x}$ with $ax+b$:
$$f:x \mapsto 3x+2(1-e^{-x})$$
Properties:

for $x\in [0, \infty)$ we have $3x \in [0, \infty)$ and
$e^x \in [1,\infty) \implies e^{-x} \in (0, 1] \implies (1-e^{-x}) \in [0,1)$
hence $f(x) \in [0, \infty)$, as required;
$f(0) = 3\cdot 0 + 2(1-\exp(-0)) = 3\cdot 0 + 2(1-1) = 0+0 = 0$, as required;
$\lim\limits_{x\to\infty} f(x)/x = \lim [3x/x+2/x - 2\exp(-x)/x] = 3+0-0 = 3$,
hence there exists $\alpha = \lim_{x\to\infty}\tfrac{f(x)}x$, which is greater than zero, as required;
for all positive $x$ we have $\exp(-x)<1$, so $\delta(x) = (1-e^{-x})>0$,
hence $f(x) = \alpha x + 2\delta(x) > \alpha x$, as required;

and it clearly is continuous, as required.
Additionally,
$$f'(x) = 3 + e^{-x} > 3 > 0$$
$$f''(x) = -e^{-x} < 0$$
so $f$ is increasing and concave (which properties were not required in the question, but concavity was mentioned in the title).
EDIT
As for the additional question from the comment:

Can we have $\lim_x (f(x)−\alpha x)=0$?

Yes, we can.
Suppose $f(x) = \alpha x+ g(x)$. Then $g(x)$ will have properties:

$g(x)>0$ for $x>0$ and $g(0) = 0$;
$\lim_x g(x) = 0$.

An example:
$$g(x) = x\cdot \exp(-x)$$
see it in WolframAlpha
This, however, is not concave – at some point $(x=1)$ it starts decreasing, but to approach its asymptote it slows down the decreasing rate at some other point $(x=2)$, which means it becomes convex above $x=2$.

