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(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{\left( {1 - \cos 2x} ight)\left( {3 + \cos x} ight)}}{{x	an 4x}}$Usin

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{\left( {1 - \cos 2x} ight)\left( {3 + \cos x} ight)}}{{x	an 4x}}$Using the identity $1 - \cos 2x = 2\sin^2 x$,the expression becomes:$\mathop {\lim }\limits_{x 	o 0} \frac{{2\sin^2 x(3 + \cos x)}}{{x 	an 4x}}$Multiply and divide by $x$ and $4x$ to use standard limits $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$ and $\mathop {\lim }\limits_{	heta 	o 0} \frac{	an 	heta}{	heta} = 1$:$= \mathop {\lim }\limits_{x 	o 0} \left( 2 \cdot \left( \frac{\sin x}{x} ight)^2 \cdot \frac{x^2}{x \cdot \frac{	an 4x}{4x} \cdot 4x} \cdot (3 + \cos x) ight)$$= \mathop {\lim }\limits_{x 	o 0} \left( 2 \cdot \left( \frac{\sin x}{x} ight)^2 \cdot \frac{1}{4 \cdot \frac{	an 4x}{4x}} \cdot (3 + \cos x) ight)$Substituting $x = 0$:$= 2 \cdot (1)^2 \cdot \frac{1}{4 \cdot 1} \cdot (3 + \cos 0)$$= 2 \cdot \frac{1}{4} \cdot (3 + 1) = 2 \cdot \frac{1}{4} \cdot 4 = 2$

$\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} = $

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