In this answer https://math.stackexchange.com/a/2101475/72031 I proved that we can sometimes avoid the limit $$\lim_{x\to 0}\frac{\sin x} {x} = 1\tag{1}$$ and instead use the simpler limit $$\lim_{x\to 0}\cos x=1\tag{2}$$ to evaluate many limits. Specifically we can use $(2)$ to show that $$\lim_{x\to 0}\frac{\sin nx} {\sin x} =n\tag{3}$$ for all rational $n$. My contention is that the above limit can not be established for irrational values of $n$ just by using $(2)$ and it necessarily requires the use of $(1)$.

Based on this thought I pose the following problem:

Let $f:\mathbb{R} \to\mathbb{R} $ be continuous with $f(0)=0$ and $$\lim_{x\to 0}\frac{f(ax)}{f(x)}=a\tag{4}$$ for all non-zero real values of $a$. Does this imply that $f'(0)$ exist and is nonzero?

I think this is false, but I could not find an easy counter-example. So either a proof or a counter-example is desired.

  • $\begingroup$ If it is continuous, doesn't proving for the rationals suffice? $\endgroup$ – Simply Beautiful Art Jan 18 '17 at 16:06
  • $\begingroup$ @SimpleArt: how does continuity of $f$ help here? $\endgroup$ – Paramanand Singh Jan 18 '17 at 16:08
  • $\begingroup$ I meant your sine limit with irrational $n$. $\endgroup$ – Simply Beautiful Art Jan 18 '17 at 16:09

I try a counter-example:

Let first be $g(x)=\sqrt{|\log (|x|)|}$. For any $a$ not zero $g(ax)-g(x)\to 0$ if $x\to 0$, $x\not =0$. (Because for $|x|$ small, $|\log |a|+\log |x||=-\log |x|-\log |a|$). Hence if we put $f(x)=x\exp(g(x))$ for $x\not =0$, and $f(0)=0$, we are done.

  • $\begingroup$ Ah, geniusly designed... $\endgroup$ – Simply Beautiful Art Jan 18 '17 at 16:43
  • $\begingroup$ Ur c’helenner vat! ... $\endgroup$ – Ewan Delanoy Jan 18 '17 at 16:49
  • $\begingroup$ I must say that your example is really very smart. +1 BTW I had a big relief as I was expecting that a counter-example exists. Thus the existence of limit of $f(ax) /f(x) $ is a much weaker hypothesis than existence of $f'(0)$. Will wait for a while before I accept any answer. $\endgroup$ – Paramanand Singh Jan 18 '17 at 16:59

However the previous answer is correct this gives a very simple one if $a $ may only be positive, then $f=|x|$ is also continuous and the limit of the ratio is also $a $ and the derivative doesn't exist in 0

  • $\begingroup$ For $a<0$, is it? $\endgroup$ – Clement C. Jan 18 '17 at 18:09
  • $\begingroup$ Already edited it $\endgroup$ – Dylan_VM Jan 18 '17 at 18:10

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