JEE Main 2013Session, Shift 1MathematicsMediumLimit, Continuity and Differentiability420× since 2002limx→0(1−cos2x)(3+cosx)xtan4x\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)} \over {x\tan 4x}}x→0limxtan4x(1−cos2x)(3+cosx) is equal toAOption A. −14- {1 \over 4}−41BOption B. 12{1 \over 2}21COption C. 1DOption D. 2CheckSkip · show solution
