limx→0xtan4x−2xtan2x(1−cos4x)2=

limx→0x(2tan2x)1−tan22x−2xtan2x(1−(1−2sin22x))2=limx→02xtan2x(11−tan22x−1)(2sin22x)2 =limx→02xtan2xtan22x1−tan22x4sin42x =limx→0xtan22x.tan4x4sin42x =14limx→0(tan2xx)2.(tan4xx)(sin2xx)4 =14×4×416 =14

limx→0xtan4x−2xtan2x(1−cos4x)2=

limx→0x(2tan2x)1−tan22x−2xtan2x(1−(1−2sin22x))2=limx→02xtan2x(11−tan22x−1)(2sin22x)2 =limx→02xtan2xtan22x1−tan22x4sin42x =limx→0xtan22x.tan4x4sin42x =14limx→0(tan2xx)2.(tan4xx)(sin2xx)4 =14×4×416 =14

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$\lim_{x \to 0} \frac{x \tan 4 x - 2 x \tan 2 x}{{(1 - \cos 4 x)}^{2}} =$

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$\frac{1}{8}$

$\frac{1}{4}$

$\frac{1}{2}$

answer is B.

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Detailed Solution

$\lim_{x \to 0} \frac{x \frac{(2 \tan 2 x)}{1 - \tan^{2} 2 x} - 2 x \tan 2 x}{{(1 - (1 - 2 \sin^{2} 2 x))}^{2}} = \lim_{x \to 0} \frac{2 x \tan 2 x (\frac{1}{1 - \tan^{2} 2 x} - 1)}{{(2 \sin^{2} 2 x)}^{2}}$

$\begin{array}{l} = \lim_{x \to 0} \frac{2 x \tan 2 x \frac{\tan^{2} 2 x}{1 - \tan^{2} 2 x}}{4 \sin^{4} 2 x} = \lim_{x \to 0} \frac{x \tan^{2} 2 x . \tan 4 x}{4 \sin^{4} 2 x} \\ = \frac{1}{4} \lim_{x \to 0} \frac{{(\frac{\tan 2 x}{x})}^{2} . (\frac{\tan 4 x}{x})}{{(\frac{\sin 2 x}{x})}^{4}} = \frac{1}{4} \times \frac{4 \times 4}{16} = \frac{1}{4} \end{array}$