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Assume $f$ is continuous. For some fixed $a\in(0,1)$ and some real number A, $$\lim\limits_{x\to0}\frac{f(x)-f(ax)}{x}=A.$$ Prove that $f$ is differentiable at $x=0$ and compute the value of $f’(0)$.

I do not know how to use the condition. Can someone help me?

Paramanand Singh
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chole
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  • If we set $g(x)=f(x)-f(ax)$ then $g$ differentiable and $g'(0)=A$. We have also $\frac{g(a^nx)}x\sim a^nA$, and $\frac{f(x)-f(a^nx)}x=\sum\limits_{k=0}^n\frac{g(a^kx)}{x}$. But I don't know how to make it rigourous to swap limits (in $x$ and in $n$) to get to $\lim_{x\to 0}\frac{f(x)-f(0)}{x}=\sum\limits_{k=0}^\infty a^k A=\frac A{1-a}$ – zwim Nov 21 '20 at 02:44
  • This is taken from the same source as https://math.stackexchange.com/q/3916455/72031 – Paramanand Singh Nov 21 '20 at 02:48
  • @zwim: please see the answer to the dupe target. – Paramanand Singh Nov 21 '20 at 02:49

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