$\ln x\approx 2\cdot \dfrac {x-1}{x+1}$, for $x\approx 1$ , How does one show this result?

Question

Where does this result come from: I saw it when viewing Andrew Woods answer on this post: What is the best way to calculate log without a calculator?.

I thought one would be able to derive it from the Taylor expansion:

$\ln \left( 1+\varepsilon \right) =\varepsilon -\dfrac {\varepsilon ^{2}}{2}$ + ...

Set $X = 1+\varepsilon $

$\ln \left( X\right) =X-1-\dfrac {\left( X-1\right) ^{2}}{2}$+...

but that does not seem to work?

Answer 1

Write $x=1+y$ so $y\ll 1$ and $$2\frac{x-1}{x+1}=\frac{y}{1+y/2}\approx y-y^2/2\approx\ln(1+y)=\ln x.$$