JEE Main 20198 Apr, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002limx→0sin2x2−1+cosx\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}x→0lim2−1+cosxsin2x equals:AOption A. 2\sqrt 22BOption B. 222 \sqrt 222COption C. 4DOption D. 424 \sqrt 242CheckSkip · show solution
