Limit of the logarithm of values very close to one: $\frac{\ln(1-\epsilon_A^2)}{\ln(1-\epsilon_B^2)}$

Question

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!

Answer 1

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 $\epsilon_A,\epsilon_B \to 0$

$$A = \frac{\ln(1-\epsilon_A^2)}{\ln(1-\epsilon_B^2)}\sim \frac{\epsilon_A^2}{\epsilon_B^2}$$

Answer 2

$\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.