The value of limx→0 (1−cos⁡2x)(3+cos⁡x)xtan⁡4x is

limx→0 (1−cos⁡2x)(3+cos⁡x)xtan⁡4x=limx→0 1−1−2sin2⁡x(3+cos⁡x)xtan⁡4x=limx→0 2sin2⁡x(3+cos⁡x)xtan⁡4x4x⋅4x=limx→0 2sin⁡xx2(3+cos⁡x)4tan⁡4x4x=24limx→0 sin⁡xx2⋅limx→0 (3+cos⁡x)limx→0 tan⁡4x4x=24⋅4=2

The value of limx→0 (1−cos⁡2x)(3+cos⁡x)xtan⁡4x is

limx→0 (1−cos⁡2x)(3+cos⁡x)xtan⁡4x=limx→0 1−1−2sin2⁡x(3+cos⁡x)xtan⁡4x=limx→0 2sin2⁡x(3+cos⁡x)xtan⁡4x4x⋅4x=limx→0 2sin⁡xx2(3+cos⁡x)4tan⁡4x4x=24limx→0 sin⁡xx2⋅limx→0 (3+cos⁡x)limx→0 tan⁡4x4x=24⋅4=2

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The value of $lim_{x \to 0} \frac{(1 - \cos 2 x) (3 + \cos x)}{xtan 4 x}$ is

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$- \frac{1}{4}$

$\frac{1}{2}$

answer is D.

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Detailed Solution

$\begin{array}{l} lim_{x \to 0} \frac{(1 - \cos 2 x) (3 + \cos x)}{xtan 4 x} \\ = lim_{x \to 0} [\frac{[1 - (1 - 2 \sin^{2} x)] (3 + \cos x)}{xtan 4 x}] \\ = lim_{x \to 0} \frac{(2 \sin^{2} x) (3 + \cos x)}{x (\frac{\tan 4 x}{4 x}) \cdot 4 x} = lim_{x \to 0} \frac{2 {(\frac{\sin x}{x})}^{2} (3 + \cos x)}{4 (\frac{\tan 4 x}{4 x})} \\ = \frac{2}{4} \frac{lim_{x \to 0} {(\frac{\sin x}{x})}^{2} \cdot lim_{x \to 0} (3 + \cos x)}{lim_{x \to 0} \frac{\tan 4 x}{4 x}} = \frac{2}{4} \cdot 4 = 2 \end{array}$