JEE Main 2014Session, Shift 1MathematicsMediumLimit, Continuity and Differentiability420× since 2002limx→0sin(πcos2x)x2\mathop {\lim }\limits_{x \to 0} {{\sin \left( {\pi {{\cos }^2}x} \right)} \over {{x^2}}}x→0limx2sin(πcos2x) is equal to :AOption A. −π- \pi−πBOption B. π\piπCOption C. π2{\pi \over 2}2πDOption D. 1CheckSkip · show solution