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

 A: The numerator is equivalent to $-\log x$. As for the denominator, rewrite it as 
$$ \log\biggl(\frac{x\log x}{\log(1-x)}\biggr) -\log x. $$
One can check the first term is $o(\log x)$, so the denominator is equivalent to $-\log x$. Thus the fraction is  equivalent to $1$, i.e. its limit, as x tends to $0_+$, is $0$.
A: 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
$$\lim_{x\to 0^{+}}\dfrac{x\cdot\dfrac{\log{x}}{\log{(1-x)}}}{\log{\left(\dfrac{\log{x}}{\log{(1-x)}}\right)}}=1.$$
A: We can proceed as follows
\begin{align}
L &= \lim_{x \to 0^{+}}\dfrac{x\cdot\dfrac{\log x}{\log(1 - x)}}{\log\left(\dfrac{\log x}{\log(1 - x)}\right)}\notag\\
&= \lim_{x \to 0^{+}}\frac{x\log x}{\log(1 - x)\log\left(\dfrac{\log x}{\log(1 - x)}\right)}\notag\\
&= \lim_{x \to 0^{+}}\log x\cdot\frac{x}{\log(1 - x)}\cdot\dfrac{1}{\log\left(\dfrac{\log x}{\log(1 - x)}\right)}\notag\\
&= -\lim_{x \to 0^{+}}\log x\cdot\dfrac{1}{\log\left(\dfrac{\log x}{\log(1 - x)}\right)}\notag\\
&= \lim_{y \to \infty}\dfrac{\log y}{\log\left(\dfrac{-\log y}{\log(1 - 1/y)}\right)}\text{ (putting }y = 1/x)\notag\\
&= \lim_{y \to \infty}\dfrac{\log y}{\log\left(y\log y\cdot\dfrac{1/y}{-\log(1 - 1/y)}\right)}\notag\\
&= \lim_{y \to \infty}\dfrac{\log y}{\log(y\log y) + \log\left(\dfrac{1/y}{-\log(1 - 1/y)}\right)}\notag\\
&= \lim_{y \to \infty}\dfrac{1}{\dfrac{\log(y\log y)}{\log y} + \dfrac{1}{\log y}\cdot\log\left(\dfrac{1/y}{-\log(1 - 1/y)}\right)}\notag\\
&= \dfrac{1}{\lim_{y \to \infty}\dfrac{\log(y\log y)}{\log y} + \lim_{y \to \infty}\dfrac{1}{\log y}\cdot\log\left(\dfrac{1/y}{-\log(1 - 1/y)}\right)}\notag\\
&= \dfrac{1}{\lim_{y \to \infty}\dfrac{\log(y\log y)}{\log y} + \lim_{y \to \infty}\dfrac{1}{\log y}\cdot\lim_{x \to 0^{+}}\log\left(\dfrac{x}{-\log(1 - x)}\right)}\notag\\
&= \dfrac{1}{\lim_{y \to \infty}\dfrac{\log(y\log y)}{\log y} + 0\cdot\log 1}\notag\\
&= \lim_{y \to \infty}\dfrac{1}{1 + \dfrac{\log\log y}{\log y}}\notag\\
&= \lim_{t \to \infty}\dfrac{1}{1 + \dfrac{\log t}{t}}\text{ (putting }t = \log y)\notag\\
&= \frac{1}{1 + 0} = 1\notag
\end{align}
We have used the following standard limits $$\lim_{x \to 0}\frac{\log(1 - x)}{x} = -1,\,\lim_{t \to \infty}\frac{\log t}{t} = 0$$ so that there is no need for L'Hospital's Rule.
