$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}$ I have to solve the limit
$$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}$$
applying Taylor's series.
$$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}=\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \frac{\sin 2x}{\cos 2x}}= \lim_{x\to 0^{+}} \frac{\ln (2 \cdot( sin x)^2)}{\ln \sin 2x - \ln \cos 2x}=    \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln \sin 2x - \ln \cos 2x}=  \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln 2 + \ln \sin x + \ln \cos x - 2\ln \cos x + 2 \ln \sin x}=  \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln 2 + 3\ln \sin x }= \lim_{x\to 0^{+}} \frac{\ln 2( sin x)^2}{\ln 2( sin x)^3}$$
$$\frac{\ln 2( sin x)^2}{\ln 2( sin x)^3} \sim \frac{\ln (2 x^2- \frac{2}{3} x^4+ o(x^4))}{\ln (2 x^3-  x^5+ o(x^5))}=  \frac{\ln (x^2)+ \ln(2 - \frac{2}{3} x^2+ o(x^2))}{\ln(x^3)+\ln (2 -  x^2+ o(x^2))} \sim \frac{2\ln x+ \ln 2}{3\ln x+\ln 2}  \sim \frac{2}{3}$$
The suggested solution in my book is $2$. can someone indicate where I made mistakes?

The OP's error has been pegged to an error with the logarithm of a difference being equated to the difference of logarithms. An alternative approach, which avoids all this, is to note that
$${\ln(1-\cos2x)\over\ln\tan2x}={\ln(2\sin^2x)\over\ln\displaystyle\left({2\tan x\over1-\tan^2x}\right)}={2\ln\sin x+\ln2\over\ln\sin x-\ln\cos x+\ln2-\ln(1-\tan^2x)}$$
Since $\ln\sin x\to-\infty$ as $x\to0^+$ while all the other terms tend to finite limits (i.e. $\ln2\to\ln2$, $\ln\cos x\to\ln\cos0=\ln1=0$, and $\ln(1-\tan^2x)\to\ln(1-\tan^20)=\ln1=0$), the requested limit is easily seen to equal $2$.
