mathop {lim }limits_{x to 0} frac{{sin (pi {{ | vedclass

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(B) We need to evaluate the limit $L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sin (\pi {{\cos }^2}x)}}{{{x^2}}}$.Since $\cos^2 x = 1 - \sin^2 x$,we

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$\pi $

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(B) We need to evaluate the limit $L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sin (\pi {{\cos }^2}x)}}{{{x^2}}}$.Since $\cos^2 x = 1 - \sin^2 x$,we can rewrite the expression as:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sin (\pi (1 - \sin^2 x))}}{{{x^2}}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sin (\pi - \pi \sin^2 x)}}{{{x^2}}}$Using the identity $\sin(\pi - 	heta) = \sin 	heta$,we get:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sin (\pi \sin^2 x)}}{{{x^2}}}$Multiply and divide by $\pi \sin^2 x$:$L = \mathop {\lim }\limits_{x 	o 0} \left( \frac{{\sin (\pi \sin^2 x)}}{{\pi \sin^2 x}} ight) 	imes \frac{{\pi \sin^2 x}}{{{x^2}}}$As $x 	o 0$,$\pi \sin^2 x 	o 0$,so $\frac{{\sin (\pi \sin^2 x)}}{{\pi \sin^2 x}} 	o 1$.Also,$\mathop {\lim }\limits_{x 	o 0} \frac{{\sin^2 x}}{{{x^2}}} = 1$.Therefore,$L = 1 	imes \pi 	imes 1 = \pi $.

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin (\pi {{\cos }^2}x)}}{{{x^2}}} = $

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