lim _{x rightarrow 0} frac{(1-cos 2 x)(3+cos | vedclass

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(D) Let $l = \lim _{x ightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x 	an 4 x}$.Using the identity $1 - \cos 2x = 2 \sin^2 x$,we have:$l = \lim _{x

Vedclass · Accepted Answer

$2$

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(D) Let $l = \lim _{x ightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x 	an 4 x}$.Using the identity $1 - \cos 2x = 2 \sin^2 x$,we have:$l = \lim _{x ightarrow 0} \frac{2 \sin^2 x (3 + \cos x)}{x 	an 4x}$.Rearranging the terms to use standard limits $\lim_{u ightarrow 0} \frac{\sin u}{u} = 1$ and $\lim_{u ightarrow 0} \frac{	an u}{u} = 1$:$l = 2 \cdot \lim _{x ightarrow 0} \left( \frac{\sin x}{x} ight)^2 \cdot \frac{x}{	an 4x} \cdot (3 + \cos x)$.Multiply and divide by $4$ to match the tangent argument:$l = 2 \cdot \lim _{x}$ ${ightarrow 0} \left( \frac{\sin x}{x} ight)^2 \cdot \frac{4x}{	an 4x} \cdot \frac{1}{4} \cdot (3 + \cos x)$.Evaluating the limits as $x ightarrow 0$:$l = 2 \cdot (1)^2 \cdot (1) \cdot \frac{1}{4} \cdot (3 + \cos 0) = 2 \cdot 1 \cdot \frac{1}{4} \cdot 4 = 2$.

$\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x} = $

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