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Let $g: (0,\infty)\rightarrow\mathbb{R}$ satisfies $\lim_{x\to 0}g(x)=0$ and $\lim_{x\to 0} \frac{g(x)-g(\frac{x}{2})}{\sqrt{x}}=1$. Show that $$\lim_{x\to 0}\frac{g(x)}{\sqrt{x}}=2+\sqrt{2}$$

Here is what I think about.

If I let $l= \lim_{x\to 0}\frac{g(x)}{\sqrt{x}}$ then I can find that $l=2+\sqrt{2}$. Because it is likely to calculate. It is not proving.

For showing this, I use definition of limit

Given $\epsilon>0,\exists \delta>0$ Such that $0<|x-0|<\delta$ And $$\left|\frac{g(x)-g(\frac{x}{2})}{\sqrt{x}}-1\right|<\epsilon$$

Then I don’t know how can I do more.Thank in advance!

LaVendEr
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  • Use the substitution $x=t^2$ and put $f(t) =g(t^2)$. After that the problem is a duplicate of https://math.stackexchange.com/q/1864058/72031 – Paramanand Singh Apr 11 '21 at 17:49

2 Answers2

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HINT: $$\lim_{x\to0} \frac{g(\frac x2)}{\sqrt{\frac x2}}=l$$ since $x/2\to 0$ as $x\to 0$

kingW3
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COMMENT.-Suppose $\lim_{x\to 0}g(x)=0$ and $\lim_{x\to 0}\dfrac{g(x)}{\sqrt x}=a\ne0$. Since $\dfrac{g(x)}{\sqrt x}$ is an indeterminate form we can apply L'Hôpital so (considering $g$ derivable) we get $\lim_{x\to 0} 2\sqrt x g'(x)=a$.

A solution of this is $\color{red}{g(x)=a\sqrt x}$ which satisfies $\lim_{x\to 0} \frac{g(x)-g(\frac{x}{2})}{\sqrt{x}}=1$. In our particular case $a=2+\sqrt2$. Is this function the only solution of this problem?

Piquito
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  • I think there may be other functions with desired property. Also it is not given if $g$ is differentiable. The problem is tricky inherently and requires a somewhat deeper analysis. Please visit the question linked in comments to this question. I have provided an answer there will all the details. – Paramanand Singh Apr 13 '21 at 02:35
  • @Paramanand Singh: Very interesting. Thank you very much. – Piquito Apr 14 '21 at 09:49