Romanian Mathematical Olympiad

a) Since $f$ is continuous and bijective, we deduce that $f(0) = 0$ and $f$ is increasing, with $\lim_{x \to \infty} f(x) = \infty$. Then $f^{-1} : [0, \infty) \to [0, \infty)$ is also increasing, with $f^{-1}(0) = 0$ and $\lim_{x \to \infty} f^{-1}(x) = \infty$. The limit from the hypothesis ensures the existence of a number $u > 0$ such that $\frac{f^{-1}(f(x)/x)}{x} > \frac{1}{2}$, for any $x > u$. Therefore, $\frac{f(x)}{x} > f(\frac{x}{2})$, for any $x > u$. Since $\lim_{x \to \infty} f(\frac{x}{2}) = \infty$, we obtain $\lim_{x \to \infty} \frac{f(x)}{x} = \infty$. Let $a > 0$ be an arbitrary fixed number.

The case $a \in (0, 1)$. There is $t > 0$ such that $f^{-1}(x) > \frac{1}{a}$, for any $x > t$. Hence
$$f^{-1}(x) > f^{-1}(ax) = f^{-1}(af(f^{-1}(x))) > f^{-1}\left(\frac{f(f^{-1}(x))}{f^{-1}(x)}\right), \text{ for any } x > t.$$
We obtain
$$\frac{f^{-1}(f(f^{-1}(x))/f^{-1}(x))}{f^{-1}(x)} < \frac{f^{-1}(ax)}{f^{-1}(x)} < 1, \text{ for any } x > t.$$
From the assumption and the condition $\lim_{x \to \infty} f^{-1}(x) = \infty$, we get
$$\lim_{x \to \infty} \frac{f^{-1}(f(f^{-1}(x))/f^{-1}(x))}{f^{-1}(x)} \underset{y=f^{-1}(x)}{=} \lim_{y \to \infty} \frac{f^{-1}(f(y)/y)}{y} = 1.$$
The case $a = 1$ is clear.

The sandwich theorem ensures $\lim_{x \to \infty} \frac{f^{-1}(ax)}{f^{-1}(x)} = 1$.

The case $a \in (1, \infty)$. Then $b = 1/a \in (0, 1)$. From the previous case, we have
$$\lim_{x \to \infty} \frac{f^{-1}(ax)}{f^{-1}(x)} = \lim_{x \to \infty} \left( \frac{f^{-1}(x)}{f^{-1}(ax)} \right)^{-1} = \lim_{x \to \infty} \left( \frac{f^{-1}(b(ax))}{f^{-1}(ax)} \right)^{-1} = 1^{-1} = 1.$$
In conclusion, $\lim_{x \to \infty} \frac{f^{-1}(ax)}{f^{-1}(x)} = 1$, for any $a > 0$.

b) Example. The bijective function $f : [0, \infty) \to [0, \infty)$, $f(x) = e^x - 1$, with the inverse $f^{-1} : [0, \infty) \to [0, \infty)$, $f^{-1}(x) = \ln(x + 1)$, satisfies the conditions from the statement.