It is clear that once we prove, that the limit exists, the value can easily be computed using L'Hospital Rule. In fact, $\lim_{x\to\infty}\frac{f'(x)}{x}=2L.$ Let me show, that $\lim\sup\frac{f'(x)}{x}\le 2L.$ The key point is the inequality, $f(x+h)\ge f(x)+hf'(x),$ for $h\ge 0.$ This immediately follows from concavity of $f$ by differentiating with respect to $h.$ Latter is equivalent to
$$\frac{f'(x)}{x}\le \frac{(x+h)^2}{xh}\left(g(x+h)-g(x)\frac{x^2}{(x+h)^2}\right)$$
for $g(x)=\frac{f(x)}{x^2},$ $x,h>0.$ Let $\varepsilon>0$ be fixed. For all sufficiently large $x>0,$
$L-\varepsilon < g(x)
$$\frac{f'(x)}{x}\le \frac{(x+h)^2}{xh}\left(L+\varepsilon- (L-\varepsilon)\frac{x^2}{(x+h)^2}\right),$$
or
$$\frac{f'(x)}{x}\le 2L+ L\frac{h}{x}+\varepsilon\frac{(x+h)^2+x^2}{xh}.$$
Now, it is easy to make $$L\frac{h}{x}+\varepsilon\frac{(x+h)^2+x^2}{xh}$$
suffitiently small by choosing $h.$ Indeed,
$$L\frac{h}{x}+\varepsilon\frac{(x+h)^2+x^2}{xh}=(L+\varepsilon)\frac{h}{x}+2\varepsilon\frac{x}{h}+2\varepsilon=2\sqrt{2\varepsilon(L+\varepsilon)}+2\varepsilon,$$
for $h=x\frac{\sqrt{2\varepsilon}}{\sqrt{L+\varepsilon}}.$ By analogy, one can prove that $\lim\inf\frac{f'(x)}{x}\ge 2L$ (just use $-h$ instead of $h$.)