How prove this limits $lim_{xto 0^{+}}frac{xcdotfrac{log{x}}{log{(1-x)}}}{log{left(frac{log{x}}{log{(1-x)}}right)}}=1$

show this limits
$$\lim_{x\to 0^{+}}\dfrac{x\cdot\dfrac{\log{x}}{\log{(1-x)}}}{\log{\left(\dfrac{\log{x}}{\log{(1-x)}}\right)}}=1$$

I fell this limits not easy to show it.

since
$$\log{(1-x)}=-x+o(x^2)\Longrightarrow x\cdot\dfrac{\log{x}}{\log{(1-x)}}\approx -\log{x}+o(\log{x})$$
and I know
$$\dfrac{\log{x}}{\log{(1-x)}}\to +\infty$$
then I don't know How to deal this problem

This Problem is from Analysis problem book exercise (MIn hui xie)

Answer

As $x\to 0^+$ we have
$$
\frac{\log x}{\log(1-x)}\sim -\frac{\log x}{x}=\frac{1}{x}\log\left(\frac{1}{x} \right)
$$
and then
$$
\frac{x\cdot\frac{\log{x}}{\log{(1-x)}}}{\log{\left(\frac{\log{x}}{\log{(1-x)}}\right)}}\sim \frac{\log(\frac{1}{x})}{\log\left(\frac{1}{x}\log(\frac{1}{x})\right)}
$$
Changing $u=\frac{1}{x}$, we have using de l'Hopital's rule
$$

\lim_{u\to\infty}\frac{\log u}{\log(u\log u)}=\lim_{u\to\infty}\frac{\log u}{\log u +1}=1.
$$
So your limit is
$$\lim_{x\to 0^{+}}\dfrac{x\cdot\dfrac{\log{x}}{\log{(1-x)}}}{\log{\left(\dfrac{\log{x}}{\log{(1-x)}}\right)}}=1.$$