JEE Main 202324 Jan, Morning shiftMathematicsEasyLimit, Continuity and Differentiability420× since 2002limt→0(11sin2t+21sin2t + ... + n1sin2t)sin2t\mathop {\lim }\limits_{t \to 0} {\left( {{1^{{1 \over {{{\sin }^2}t}}}} + {2^{{1 \over {{{\sin }^2}t}}}}\, + \,...\, + \,{n^{{1 \over {{{\sin }^2}t}}}}} \right)^{{{\sin }^2}t}}t→0lim(1sin2t1+2sin2t1+...+nsin2t1)sin2t is equal toAOption A. n(n+1)2{{n(n + 1)} \over 2}2n(n+1)BOption B. nCOption C. n2^22 + nDOption D. n2^22CheckSkip · show solution