JEE Main 202131 Aug, Morning shiftMathematicsEasyLimit, Continuity and Differentiability420× since 2002limx→0sin2(πcos4x)x4\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( {\pi {{\cos }^4}x} \right)} \over {{x^4}}}x→0limx4sin2(πcos4x) is equal to :AOption A. π2{\pi ^2}π2BOption B. 2π22{\pi ^2}2π2COption C. 4π24{\pi ^2}4π2DOption D. 4π4\pi4πCheckSkip · show solution
