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

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(B) We are given the limit $L = \lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}$.Using the identity $\cos ^2 x = 1 - \sin ^2 x$,we

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$\pi$

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(B) We are given the limit $L = \lim _{x ightarrow 0} \frac{\sin \left(\pi \cos ^2 xight)}{x^2}$.Using the identity $\cos ^2 x = 1 - \sin ^2 x$,we can rewrite the expression as:$L = \lim _{x ightarrow 0} \frac{\sin \left(\pi(1 - \sin ^2 x)ight)}{x^2}$$L = \lim _{x ightarrow 0} \frac{\sin \left(\pi - \pi \sin ^2 xight)}{x^2}$Since $\sin(\pi - 	heta) = \sin 	heta$,we have:$L = \lim _{x ightarrow 0} \frac{\sin \left(\pi \sin ^2 xight)}{x^2}$Now,multiply and divide by $\pi \sin ^2 x$:$L = \lim _{x}$ ${ightarrow 0} \left( \frac{\sin \left(\pi \sin ^2 xight)}{\pi \sin ^2 x} 	imes \frac{\pi \sin ^2 x}{x^2} ight)$Using the standard limit $\lim _{	heta ightarrow 0} \frac{\sin 	heta}{	heta} = 1$ and $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$:$L = 1 	imes \pi 	imes (1)^2 = \pi$.

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

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