JEE Main 201610 Apr, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002limx→0 (1−cos2x)22x tanx −xtan2x\mathop {\lim }\limits_{x \to 0} \,{{{{\left( {1 - \cos 2x} \right)}^2}} \over {2x\,\tan x\, - x\tan 2x}}x→0lim2xtanx−xtan2x(1−cos2x)2 is :AOption A. −-− 2BOption B. −-− 12{1 \over 2}21COption C. 12{1 \over 2}21DOption D. 2CheckSkip · show solution
