<s>[INST] $\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}$ 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?
 [/INST] I think the following step is wrong:$$\lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln \sin x}{\ln \sin 2x - \ln \cos 2x}=  \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln \sin x}{\ln 2 + \ln \sin x + \ln \cos x - 2\ln \cos x + 2 \ln \sin x}.$$
I think you have used the false identity $$ \ln\cos2x \equiv 2\ln\cos x - 2\ln\sin x. $$
It is false because $$ \ln\cos 2x = \ln(\cos^2x-\sin^2x) \neq\ln\cos^2x - \ln\sin^2x = 2\ln\cos x - 2\ln\sin^2x. $$
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