lim _{x rightarrow 0} frac{sin ^{2}left(pi co | vedclass

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(C) We need to evaluate $L = \lim _{x \rightarrow 0} \frac{\sin ^{2}\left(\pi \cos ^{4} x\right)}{x^{4}}$.Since $\cos x \rightarrow 1$ as $x \rightarr

Vedclass · Accepted Answer

$4 \pi^{2}$

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(C) We need to evaluate $L = \lim _{x ightarrow 0} \frac{\sin ^{2}\left(\pi \cos ^{4} xight)}{x^{4}}$.Since $\cos x ightarrow 1$ as $x ightarrow 0$,the argument $\pi \cos^4 x ightarrow \pi$.Let $u = \pi \cos^4 x$. Then $\sin^2(u) = \sin^2(\pi - u) = \sin^2(\pi(1 - \cos^4 x))$.As $x ightarrow 0$,$1 - \cos^4 x = (1 - \cos^2 x)(1 + \cos^2 x) = \sin^2 x (1 + \cos^2 x)$.So,$L = \lim _{x ightarrow 0} \frac{\sin^2(\pi \sin^2 x (1 + \cos^2 x))}{x^4}$.Using the limit $\lim_{	heta ightarrow 0} \frac{\sin 	heta}{	heta} = 1$,we have:$L = \lim _{x}$ ${ightarrow 0} \left[ \frac{\sin(\pi \sin^2 x (1 + \cos^2 x))}{\pi \sin^2 x (1 + \cos^2 x)} ight]^2 \cdot \frac{\pi^2 \sin^4 x (1 + \cos^2 x)^2}{x^4}$.$L = 1^2 \cdot \pi^2 \cdot \lim_{x ightarrow 0} \left( \frac{\sin x}{x} ight)^4 \cdot (1 + \cos^2 x)^2$.$L = \pi^2 \cdot (1)^4 \cdot (1 + 1)^2 = \pi^2 \cdot 4 = 4 \pi^2$.

$\lim _{x \rightarrow 0} \frac{\sin ^{2}\left(\pi \cos ^{4} x\right)}{x^{4}}$ is equal to :

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