Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
How can I prove Infinitesimal Limit

Let $$\lim_{x\to 0}f(x)=0$$ and $$\lim_{x\to 0}\frac{f(2x)-f(x)}{x}=0$$ Then what is the value of $$\lim_{x\to 0}\frac{f(x)}{x}$$

share|cite|improve this question

marked as duplicate by David Mitra, Henry T. Horton, Fabian, Nameless, Davide Giraudo Dec 30 '12 at 18:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

What have you tried yourself? – Eckhard Dec 30 '12 at 16:48
I think that the above value depend on the derivation of $f$! – aliakbar Dec 30 '12 at 16:54
See here. – David Mitra Dec 30 '12 at 18:01

2 Answers 2

For every $\delta > 0$ there exists an $\varepsilon > 0$ such that $|f(2x) - f(x)| < \delta\, |x|$ if $x\neq 0$ and $|x| < \varepsilon$. Then for all integers $k \geq 0$ $$ \begin{eqnarray} |f(2x) - f(2^{-k}x)| &=& |f(2x) - f(x) + f(x) - f(2^{-1}x) + \dotsc + f(2^{1-k}x) - f(2^{-k}x)|\\ &<& \delta \,|x| + \delta\, 2^{-1}|x| + \dotsc + \delta \, 2^{-k}|x|\\ &<& 2\delta \, |x| \end{eqnarray}$$ and since $\lim_{k \to \infty}f(2^{-k}x) = 0$ this implies that $|f(2x)| \leq 2\delta \, |x|$ if $|x| < \varepsilon$. Therefore $$\lim_{x\to 0} \frac{f(x)}{x} = 0.$$

share|cite|improve this answer

If $\displaystyle\lim_{x\to0}\frac{f(x)}{x}=L$ then $$ \lim_{x\to0}\frac{f(2x)}{x} = 2\lim_{x\to0}\frac{f(2x)}{2x} = 2\lim_{u\to0}\frac{f(u)}{u} = 2L. $$ Then $$ \lim_{x\to0}\frac{f(2x)-f(x)}{x} = \lim_{x\to0}\frac{f(2x)}{x} - \lim_{x\to0}\frac{f(x)}{x} =\cdots $$ etc.

Later note: What is written above holds in cases in which $\displaystyle\lim_{x\to0}\frac{f(x)}{x}$ exists. The question remains: If both $\lim_{x\to0}f(x)=0$ and $\lim_{x\to0}(f(2x)-f(x))/x=0$ then does it follow that $\displaystyle\lim_{x\to0}\frac{f(x)}{x}$ exists?

share|cite|improve this answer
How do you know the limit exists? (-1) for that. – Nameless Dec 30 '12 at 16:54
We do not know that $f$ is derivative! – aliakbar Dec 30 '12 at 16:56
@aliakbar You mean differentiable? If $f$ is continuous at $0$ then the existence of that limit is equivalent to differentiability of $f$ at $0$. Of course, we don't even know if $f$ is continuous at $0$ let alone differentiable – Nameless Dec 30 '12 at 17:06
@Nameless : I started from the assumption that the limit exists, so what I wrote applies in all cases in which the limit exists. I believe I can show there are some functions $f$ for which $\lim_{x\to0}f(x)=0$ but $\lim_{x\to0}(f(x)/x)$ does not exist. – Michael Hardy Dec 30 '12 at 17:14
@MichaelHardy If you can construct such an example and post it here I will upvote your answer. (you also want $\frac{f(2x)-f(x)}{x}\to 0$) – Nameless Dec 30 '12 at 17:16

Not the answer you're looking for? Browse other questions tagged or ask your own question.