Limit of the logarithm of values very close to one: $\frac{\ln(1-\epsilon_A^2)}{\ln(1-\epsilon_B^2)}$ I am thinking of a way to obtain an approximation I found in a paper I'm currently reading:
We need an approximation of the logarithm taken in a value very very close to 1. Lets say:
$A = \frac{\ln(1-\epsilon_A^2)}{\ln(1-\epsilon_B^2)}$
and we basically want to take the limit when $\epsilon_B\to 0.$
It happens that the paper claims that this is:  $ A \approx \frac{\epsilon_A^2}{\epsilon_B^2}$.
Does anybody have some idea of how this result popped up?
Any clue on logarithm rules or limits..
Thanks!
 A: Recall that for $x\to 0$ 
$$\frac{\ln (1+x)}{x}\to 1\implies \ln (1+x)\sim x$$
indeed since for $x\to 0$ 
$$\left(1+x\right)^\frac1x\to e \implies \log (1+x)^\frac1x\to\log e=1$$
therefore in this case for $\epsilon_A,\epsilon_B \to 0$
$$A = \frac{\ln(1-\epsilon_A^2)}{\ln(1-\epsilon_B^2)}\sim \frac{\epsilon_A^2}{\epsilon_B^2}$$
A: $\ln(1-x)
=\int_1^{1-x}\dfrac{dt}{t}
$
so
$-\ln(1-x)
=-\int_1^{1-x}\dfrac{dt}{t}
=\int^1_{1-x}\dfrac{dt}{t}
=\int^0_{-x}\dfrac{dt}{1+t}
=\int_0^{x}\dfrac{dt}{1-t}
$
so
$-\ln(1-x)-x
=\int_0^{x}\dfrac{dt}{1-t}-x
=\int_0^{x}(\dfrac{1}{1-t}-1)dt
=\int_0^{x}\dfrac{t\,dt}{1-t}
$
so,
if $1 > x > 0$,
$-\ln(1-x)-x
\gt 0$
and
$-\ln(1-x)-x
\lt \int_0^{x}\dfrac{t\,dt}{1-x}
=\dfrac{x^2}{2(1-x)}
$.
In particular,
if $0 < x < \frac12$,
$0 < -\ln(1-x)-x
\lt x^2$.
Of course
the power series is
$-\ln(1-x)
= x+\dfrac{x^2}{2}+...
$.
We can get the
first two terms by
$\dfrac{x^2}{2(1-x)}
=\dfrac{x^2-x^3+x^3}{2(1-x)}
=\dfrac{x^2(1-x)+x^3}{2(1-x)}
=\dfrac{x^2}{2}+\dfrac{x^3}{2(1-x)}
$.
This can be readily extended
to get the full power series
for $\ln(1-x)$
using
$\dfrac{1-x^n}{1-x}
=\sum_{k=0}^{n-1} x^k
$
so that
$\dfrac{1}{1-x}
=\sum_{k=0}^{n-1} x^k
+\dfrac{x^n}{1-x}
$
and integrating.
