Prove that $ \lim_{x\to 0}\bigg[ \ln \bigg(\frac{1+\cos(x)}{1-\cos(x)} \bigg)\bigg] \sim 2\ln |x|$

Question

Hello may someone please help me prove this relation

\begin{equation} \lim_{x\to 0}\bigg[ \ln \bigg(\frac{1+\cos(x)}{1-\cos(x)} \bigg)\bigg] \sim 2\ln |x|. \end{equation}

My understanding so far is to manipulate it so that the left-hand side is equal to

\begin{equation} \lim_{x\to 0}\bigg[ 2\ln\bigg( \frac{1+\cos (x)}{\sin (x)} \bigg) \bigg] . \end{equation}

Thanks in advance!

Answer 1

$$\log\frac{1+\cos x}{1-\cos x}=\log\frac{(1+\cos x)^2}{\sin^2 x}=2\log(1+\cos x)-2\log|\sin x|.$$

The first term is constant, while the sine is asymptotic to $|x|$ (for $x\to0$).

Answer 2

I think your fraction's upside down:

$$\ln\tfrac{1-\cos x}{1+\cos x}=\ln\tan^2\tfrac{x}{2}=2\ln|\tan\tfrac{x}{2}|\approx2\ln|x|-2\ln 2\sim2\ln|x|\stackrel{x\to0^+}{\to}-\infty.$$