lim _{x rightarrow 0} frac{1-cos x cos 2 x}{s | vedclass

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(B) We need to evaluate the limit $L = \lim _{x \rightarrow 0} \frac{1-\cos x \cos 2 x}{\sin ^2 x}$.Using the identity $\cos 2x = 1 - 2\sin^2 x$,we re

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$\frac{5}{2}$

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(B) We need to evaluate the limit $L = \lim _{x \rightarrow 0} \frac{1-\cos x \cos 2 x}{\sin ^2 x}$.Using the identity $\cos 2x = 1 - 2\sin^2 x$,we rewrite the numerator:$1 - \cos x(1 - 2\sin^2 x) = 1 - \cos x + 2\sin^2 x \cos x$.So,$L = \lim _{x \rightarrow 0} \frac{1 - \cos x + 2\sin^2 x \cos x}{\sin^2 x}$.$L = \lim _{x \rightarrow 0} \left( \frac{1 - \cos x}{\sin^2 x} + \frac{2\sin^2 x \cos x}{\sin^2 x} \right)$.Using $\sin^2 x = 1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$:$L = \lim _{x \rightarrow 0} \left( \frac{1 - \cos x}{(1 - \cos x)(1 + \cos x)} + 2\cos x \right)$.$L = \lim _{x \rightarrow 0} \left( \frac{1}{1 + \cos x} + 2\cos x \right)$.Substituting $x = 0$:$L = \frac{1}{1 + 1} + 2(1) = \frac{1}{2} + 2 = \frac{5}{2}$.

$\lim _{x \rightarrow 0} \frac{1-\cos x \cos 2 x}{\sin ^2 x} = $

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