I have issues showing the following fact: For $x \in (0, 1/2)$, $$\frac{\log(2/x)}{\log(1/(1-x)} \approx \frac{1}{x}$$
I have tried with the trick of adding $1-1$ in the log, and using Taylor expansion of $\log(1-x) \approx x$, but I get:
$\frac{1+\frac{2}{x}}{1+\frac{1}{1-x}}= (\frac{x+2}{x})(\frac{1-x}{2-x})=\frac{-x^2+x+2}{-x^2+2x}$, which is not what I expect.
Are there any other tricks to get what I think is correct? Thanks.
Numerator $log(2/x)=log(2/(x-2+2))=log(1/(x/2-1+1))=-log(1+(x/2-1)\approx 1-x/2$
Denominator $log(1/(1-x))=-log(1-x)\approx x$
Ratio $\approx(1-x/2)/x$.
Major problem: for $x$ near $0$, numerator appprox. is not very good.