# How can I prove Infinitesimal Limit

how can I prove this trouble: 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 $$\lim_{x\to 0} \frac{f(x)}{x} = 0$$

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It's not clear to me why did you write $$\lim\limits_{x\to 0}\frac{f(2x)-f(x)}{x} = \lim\limits_{x\to 0}\frac{f(x)+o(x)}{x}$$ and where do you use that $f(x) = o(1)$? –  Ilya Dec 8 '11 at 10:02
surely I'm wrong .... but I replaced$${f(2x)-f(x)}=o({x})\Rightarrow {f(2x)}=f(x)+o({x})$$ –  FrConnection Dec 8 '11 at 10:09
ok , but my conclusion are right? or i'm in error? –  FrConnection Dec 8 '11 at 10:25
I thinks after the word "Well" solution is incorrect - and I pointed you the reasons in my first comment –  Ilya Dec 8 '11 at 12:57
can you show me the right way? –  FrConnection Dec 8 '11 at 13:19

As pointed out by Ilya the argument of the OP is incorrect. Here is my proposal:

There is a function $x\to g(x)$ with $g(x)\to 0$ $\ (x\to 0)$ such that

$$f(2x)-f(x)=x\ g(x)\ .$$

As $f(x)\to 0$ $\ (x\to 0)$ we can set up the following telescopic series:

$$f(x)\ =\ \sum_{k=1}^\infty \bigl(f(2\cdot 2^{-k}x)- f(2^{-k}x)\bigr) \ =\ x\ \sum_{k=1}^\infty 2^{-k} g(2^{-k}x)\ .$$

Since $\sum_{k\geq 1} 2^{-k}=1$ it follows that

$$\left|{f(x)\over x}\right|\ \leq\ \sup_{0<|t|<|x|}\ |g(t)|\to 0\qquad (x\to 0)\ .$$

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Cool argument! ${}{}{}{}{}$ –  Jonas Teuwen Dec 8 '11 at 16:34
Agree with Jonas, a very nice solution –  Ilya Dec 8 '11 at 16:38

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 $x<\delta$.

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