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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 $$\lim_{x\to 0^+} f(x) .$$

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Assume $|f(x)|<M$ for all $x\in[0,1]$. Let $\epsilon>0$ be given. Then we can find $n\in\mathbb N$ such that $b^n\epsilon>M$ (here we use that $b>1$). For $0<x<1/a^n$ we find by induction that $f(a^nx)=b^nf(x)$ and hence $|f(x)|<\epsilon$. This expresses that $$\lim_{x\to0^+}f(x)=0.$$ Note that we did not really make use of $a>1$. Also, $f$ need not be continuous anywhere.

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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 $b>1$.

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