\(\displaystyle\lim_{x\to\frac{\pi}{4}}\;(\tan2x.\ln\tan x)=\lim_{x\to\frac{\pi}{4}}\;\frac{\ln\tan x}{\cot2x}\left(=\frac00\right)\overset{H}{=}\lim_{x\to\frac{\pi}{4}}\;\frac{\frac{1}{\tan x}.\frac{1}{\cos^2x}}{-\frac{1}{\sin^22x}.2}\\=-\displaystyle\lim_{x\to\frac{\pi}{4}}\;\frac{\frac{1}{\sin x}.\frac{1}{\cos x}}{\frac{2}{\sin^22x}}=-\lim_{x\to\frac{\pi}{4}}\;\frac{\sin^22x}{2.\sin x.\cos x}=-\lim_{x\to\frac{\pi}{4}}\;\frac{\sin^22x}{\sin2x}\\=-\displaystyle\lim_{x\to\frac{\pi}{4}}\;\sin2x=-\sin\frac{\pi}{2}=-1\)