JEE Main 2016Session, Shift 1MathematicsEasyLimit, Continuity and Differentiability420× since 2002Let p=limx→0+(1+tan2x)12xp = \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 + {{\tan }^2}\sqrt x } \right)^{{1 \over {2x}}}}p=x→0+lim(1+tan2x)2x1 then logloglog ppp is equal to :AOption A. 12{1 \over 2}21BOption B. 14{1 \over 4}41COption C. 222DOption D. 111CheckSkip · show solution