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

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(C) We need to evaluate the limit: $\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}$.Using the trigonometric identity $\sin(\pi -

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$\pi$

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(C) We need to evaluate the limit: $\lim _{x ightarrow 0} \frac{\sin \left(\pi \cos ^2 xight)}{x^2}$.Using the trigonometric identity $\sin(\pi - 	heta) = \sin 	heta$,we can rewrite the numerator:$\sin(\pi \cos^2 x) = \sin(\pi - \pi \cos^2 x) = \sin(\pi(1 - \cos^2 x)) = \sin(\pi \sin^2 x)$.Now the limit becomes: $\lim _{x ightarrow 0} \frac{\sin(\pi \sin^2 x)}{x^2}$.Multiply and divide by $\pi \sin^2 x$:$= \lim _{x ightarrow 0} \left( \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} 	imes \frac{\pi \sin^2 x}{x^2} ight)$.Since $\lim _{	heta ightarrow 0} \frac{\sin 	heta}{	heta} = 1$ and $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$,we have:$= 1 	imes \pi 	imes \lim _{x ightarrow 0} \left( \frac{\sin x}{x} ight)^2 = 1 	imes \pi 	imes 1^2 = \pi$.

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

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