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

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(B) We need to evaluate the limit: $\mathop {\lim}\limits_{x 	o 0} \frac{{\sin \left( {\pi {{\cos }^2}x} ight)}}{{{x^2}}}$Using the identity $\cos^

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$\pi$

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(B) We need to evaluate the limit: $\mathop {\lim}\limits_{x 	o 0} \frac{{\sin \left( {\pi {{\cos }^2}x} ight)}}{{{x^2}}}$Using the identity $\cos^2 x = 1 - \sin^2 x$,we have:$\mathop {\lim}\limits_{x 	o 0} \frac{{\sin \left( {\pi (1 - \sin^2 x)} ight)}}{{{x^2}}}$$= \mathop {\lim}\limits_{x 	o 0} \frac{{\sin \left( {\pi - \pi \sin^2 x} ight)}}{{{x^2}}}$Since $\sin(\pi - 	heta) = \sin 	heta$,the expression becomes:$= \mathop {\lim}\limits_{x 	o 0} \frac{{\sin \left( {\pi \sin^2 x} ight)}}{{{x^2}}}$Multiply and divide by $\pi \sin^2 x$:$= \mathop {\lim}\limits_{x 	o 0} \left( \frac{{\sin \left( {\pi \sin^2 x} ight)}}{{\pi \sin^2 x}} \cdot \frac{{\pi \sin^2 x}}{{{x^2}}} ight)$$= \left( \mathop {\lim}\limits_{x 	o 0} \frac{{\sin \left( {\pi \sin^2 x} ight)}}{{\pi \sin^2 x}} ight) \cdot \pi \cdot \left( \mathop {\lim}\limits_{x 	o 0} \frac{{\sin x}}{x} ight)^2$Using the standard limit $\mathop {\lim}\limits_{	heta 	o 0} \frac{{\sin 	heta}}{	heta} = 1$:$= 1 \cdot \pi \cdot (1)^2 = \pi$

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

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