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

For x≠0,(1-cos2x)(3+cosx)xtan4x=2sin2xx2×(3+cosx)4tan4x4xTherefore, the required limit =2×1×44×1=2

For x≠0,(1-cos2x)(3+cosx)xtan4x=2sin2xx2×(3+cosx)4tan4x4xTherefore, the required limit =2×1×44×1=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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For $x \neq 0, \frac{(1 - \cos 2 x) (3 + \cos x)}{x \tan 4 x}$

$= \frac{2 \frac{\sin^{2} x}{x^{2}} \times (3 + \cos x)}{4 \frac{\tan 4 x}{4 x}}$

Therefore, the required limit $= \frac{2 \times 1 \times 4}{4 \times 1} = 2$

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