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

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(B) Given the limit: $\lim _{x \rightarrow 0} \frac{\sin (\pi \cos ^2 x)}{x^2}$ Since $\cos ^2 x = 1 - \sin ^2 x$,we have $\pi \cos ^2 x = \pi - \pi \

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$\pi$

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(B) Given the limit: $\lim _{x ightarrow 0} \frac{\sin (\pi \cos ^2 x)}{x^2}$ Since $\cos ^2 x = 1 - \sin ^2 x$,we have $\pi \cos ^2 x = \pi - \pi \sin ^2 x$. Using the identity $\sin (\pi - 	heta) = \sin 	heta$,we get $\sin (\pi - \pi \sin ^2 x) = \sin (\pi \sin ^2 x)$. Now,the limit becomes: $\lim _{x ightarrow 0} \frac{\sin (\pi \sin ^2 x)}{x^2}$. Multiplying and dividing by $\pi \sin ^2 x$: $\lim _{x}$ ${ightarrow 0} \left( \frac{\sin (\pi \sin ^2 x)}{\pi \sin ^2 x} ight) 	imes \left( \frac{\pi \sin ^2 x}{x^2} ight)$. As $x ightarrow 0$,$\sin ^2 x ightarrow 0$,so $\frac{\sin (\pi \sin ^2 x)}{\pi \sin ^2 x} ightarrow 1$. Also,$\lim _{x ightarrow 0} \frac{\sin ^2 x}{x^2} = 1$. Therefore,the limit is $1 	imes \pi 	imes 1 = \pi$.

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

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