75th 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 $$\underset{x\to \infty }{\lim }\frac{f^{-1}(f(f^{-1}(x))/f^{-1}(x))}{f^{-1}(x)}\underset{y=f^{-1}(x)}{=}\underset{y\to \infty }{\lim }\frac{f^{-1}(f(y)/y)}{y}=1.$$ 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 $$\underset{x\to \infty }{\lim }\frac{f^{-1}(ax)}{f^{-1}(x)}=\underset{x\to \infty }{\lim }{\left(\frac{f^{-1}(x)}{f^{-1}(ax)}\right)}^{-1}=\underset{x\to \infty }{\lim }{\left(\frac{f^{-1}(b(ax))}{f^{-1}(ax)}\right)}^{-1}=1^{-1}=1.$$ 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.