Show that $\log\left(\log\left(\frac{1}{x}\right)\right)=o(\log(x))$ as $x\rightarrow0^+$ using a Taylor expansion

Question

I need to show that $\log\left(\log\left(\frac{1}{x}\right)\right)=o(\log(x))$ as $x\to0^+$.

One way to show it is to use L'Hopital's rule

$$\lim_{x\to 0^+} \frac{\log\left(\log\left(\frac{1}{x}\right)\right)}{\log(x)}=\lim_{x\to 0^+} \frac{\log\left(\frac{1}{x}\right)^{-1}x(-x^{-2})}{x^{-1}}=\lim_{x\to 0^+} \frac{1}{\log(x)}=0$$

I have tried to show this result using a Taylor series expansion of $\log\left(\log\left(\frac{1}{x}\right)\right)$ around $x=0$ but I'm finding it difficult. Could I have some help.

Answer 1

We know that $\displaystyle{\lim_{t\to+\infty}\frac{\ln(t)}{t}=0}$.

Since $\displaystyle{\lim_{x\to0^+}\ln(1/x)=+\infty}$, we have $\displaystyle{\lim_{x\to0^+}\frac{\ln(\ln(1/x))}{\ln(1/x)}=0}$, hence the conclusion.

Remark

The OP asked to show this using Taylor expansion of $\ln(\ln(x))$ around $x=0$. This isn't possible ! Functions having a Taylor expansion around $x=a$ are (among other things) locally bounded in the neighborhood of $a$.