Function satisfying $\lim_{x\to 0}\frac{f(ax)}{f(x)}=a$

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.