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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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    I deleted one comment not to overload the discussion. I don't know the solution, one idea is that if $\lim\frac{f(x)}{x}$ exists then it is zero since $f(2x) = f(x)+o(x)$, though I haven't proved the existence.2011-12-08

2 Answers 2

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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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    Agree with Jonas, a very nice solution2011-12-08
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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 $x<\delta$.