let $f$ a continuos function on $\mathbb{R}$
such that:
$\lim_{x \to 0} \frac{f(2x)-f(x)}{x} =l$
proof that $f$ is derivable at 0 and $f'(0)=l$
i have tried to use the definition but i couldn't extract $f(2x)$
please help me with this question
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1The title and the question differ very much. What is the hypothesis and what is the conclusion? – geetha290krm Nov 20 '22 at 12:03
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1See link – Abel Wong Nov 20 '22 at 12:11
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1Notice that : $$\dfrac{f(x) - f(0)}{x} = \sum_{n = 0}^{+\infty} \dfrac{f(x/2^n) - f(x/2^{n + 1})}{x/2^{n + 1}} \dfrac{1}{2^{n + 1}}$$ – Essaidi Nov 20 '22 at 12:44
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The definition of $f$ being derivable at $0$ is $$\lim_{x\to 0}\frac{f(x)-f(0)}{x}.$$ You can split $$\frac{f(2x)-f(x)}{x}=2\frac{f(2x)-f(0)}{2x}-\frac{f(x)-f(0)}{x}.$$ Can you follow from here?
Edit: as requested, I'm developing the proof a bit more. Take limits in the expression above to get $$l=\lim_{x\to 0}\left( 2\frac{f(2x)-f(0)}{2x}-\frac{f(x)-f(0)}{x}\right).$$ Now, use the fact that the limit of the sum is the sum of the limits iff the summands' limits exist. So, $$l=2\lim_{x\to 0}\frac{f(2x)-f(0)}{2x}-\lim_{x\to 0}\frac{f(x)-f(0)}{x}$$ iff both of the summands' limits exist. But, since $f$ is continuous (particularly at $0$) it doen't matter in which way you take the limit $\lim_{x\to 0}f(x)$. Therefore, these limits are equal if they exist: $$\lim_{x\to 0}\frac{f(2x)-f(0)}{2x}=\lim_{x\to 0}\frac{f(x)-f(0)}{x}.$$ Then, $$l=\lim_{x\to 0}\frac{f(x)-f(0)}{x}$$ iff the limit in the RHS exists. Done.
Little Jonny
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@Andrés Cárcel can you continue on your proof because i still don't get the idea – user912835 Nov 20 '22 at 12:05
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This answer is based on the assumption that the limit exists. You have not yet proven the limit exists. And proof f is derivable at 0(i.e. limit exists) is one part of the question. – Abel Wong Nov 20 '22 at 13:02
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