If $f(x)$ and $(f(2x)-f(x))/x$ have limit $0$ as $x\to 0$, then $f(x)/x\to 0$ 
Prove that if $\lim_{x\to 0} f(x) = 0$ and
  $\lim_{x\to 0} \frac{f(2x)-f(x)}{x}= 0$,
  then
  $\lim_{x\to 0} \frac{f(x)}{x} = 0.$

I try to solve it in this way:
$f(x)$ is infinitesimal because $\lim_{x\to 0} f(x) = 0,$ 
$\lim_{x\to 0} \frac{f(2x)-f(x)}{x}= 0,\Rightarrow {f(2x)-f(x)}=o({x})\Rightarrow  {f(2x)}=f(x)+o({x}).$
Well
$$\lim_{x\to 0} \frac{f(2x)-f(x)}{x}= \lim_{x\to 0} \frac{f(x)+o({x})}{x}= \lim_{x\to 0} \frac{f(x)}{x}+\lim_{x\to 0}\frac{o({x})}{x}=0$$
$\lim_{x\to 0}\frac{o({x})}{x}=0$, of course; then
$$\lim_{x\to 0} \frac{f(x)}{x} = 0$$

This is Problem 3.2.7 from Problems in real analysis by Teodora-Liliana T. Rădulescu, Vincentiu D. Radulescu, Titu Andreescu; p.121-122. The only difference is that they use  $\lim\limits_{x\to 0} \frac{f(x)-f(x/2)}x=0$, which is obviously equivalent to the condition from the question, and they work with $x>0$, which can be easily modified.  I'll give a sketch of their solution here.
The basic idea is to notice $\lim\limits_{x\to 0} \frac{f(x)-f(x/2)}x=0$ implies that for given $\varepsilon>0$ there is $\delta>0$ such that
$$|f(x)-f(x/2)|<\varepsilon |x|$$
whenever  $x<\delta$. For any fixed $x<\delta$ we have
$|f(x/2^n)-f(x/2^{n+1})|<\varepsilon |x|/2^n$ and by triangle inequality we get
$$|f(x)-f(x/2^n)| \le 2\varepsilon |x|.$$
Now taking the limit $n\to\infty$ and using $\lim\limits_{x\to 0} f(x)=0$ gives
$$|f(x)|\le 2\varepsilon |x|$$
$$\frac{|f(x)|}{|x|} \le 2\varepsilon$$
for any $x<\delta$.
