limx→0(1−cos2x)(3+cosx)xtan4x is equal to

Ltx→0(1−cos2x)x2.(3+cosx)tan4xx=(4)244=2

Ltx→0(1−cos2x)x2.(3+cosx)tan4xx=(4)244=2

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$\lim_{x \to 0} \frac{(1 - \cos 2 x) (3 + \cos x)}{x \tan 4 x}$ is equal to

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$- \frac{1}{4}$

$\frac{1}{2}$

answer is D.

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$\underset{x \to 0}{L t} \frac{\frac{(1 - \cos 2 x)}{x^{2}} . (3 + \cos x)}{\frac{\tan 4 x}{x}} = \frac{\frac{(4)}{2} 4}{4} = 2$

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