What is the answer of this problem? Suppose that $f(x)$ is bounded on interval $[0,1]$, and for $0 < x < 1/a$, we have $f(ax)=bf(x)$. (Note that $a, b>1$). Please calculate $$\lim_{x\to 0^+} f(x) .$$

We get $f\left(a^{n}x\right)=b^{n}f\left(x\right)$ for $0<x<a^{-n}$.
If the limit does not equal $0$ then there is a series with $x_{n}\in(0,a^{-n})$
 with $f\left(x_{n}\right)\geq\varepsilon>0$
for each $n$. Then $f\left(a^{n}x_{n}\right)=b^{n}f\left(x_{n}\right)\geq b^{n}\varepsilon$.
This contradicts the boundedness of $f$ since $b>1$.
