limx→0sin(π.cos2x)x2 is equal to

limx→0sin(πcos2x)x2 =limx→0sin(π(1−sin2x))x2 =limx→0sin(πsin2x)(πsin2x).πsin2xx2 =1×π =π

limx→0sin(π.cos2x)x2 is equal to

limx→0sin(πcos2x)x2 =limx→0sin(π(1−sin2x))x2 =limx→0sin(πsin2x)(πsin2x).πsin2xx2 =1×π =π

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$\lim_{x \to 0} \frac{\sin (π . \cos^{2} x)}{x^{2}}$ is equal to

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$π$

$\frac{π}{2}$

$- π$

answer is D.

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Detailed Solution

$\lim_{x \to 0} \frac{\sin (π \cos^{2} x)}{x^{2}} = \lim_{x \to 0} \frac{\sin (π (1 - \sin^{2} x))}{x^{2}} = \lim_{x \to 0} \frac{\sin (π \sin^{2} x)}{(π \sin^{2} x)} . \frac{π \sin^{2} x}{x^{2}} = 1 \times π = π$

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