JEE Main 201815 Apr, Evening shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002limx→0xtan2x−2xtanx(1−cos2x)2\mathop {\lim }\limits_{x \to 0} {{x\tan 2x - 2x\tan x} \over {{{\left( {1 - \cos 2x} \right)}^2}}}x→0lim(1−cos2x)2xtan2x−2xtanx equals :AOption A. 14{1 \over 4}41BOption B. 1COption C. 12{1 \over 2}21DOption D. −-− 12{1 \over 2}21CheckSkip · show solution