JEE Main 2018Limits, Continuity & DifferentiabilityLimits, Continuity and DifferentiabilityMediumQuestionlimt→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 toOptionsAn(n+1)2{{n(n + 1)} \over 2}2n(n+1)BnCn2^22 + nDn2^22Check AnswerHide SolutionSolutionlimt→0(1cosec2t+2cosec2t+……..+ncosec2t)sin2t=limt→0n((1n)cosec2t+(2n)cosec2t+……..+1)sin2t=n\begin{aligned} & \lim _{t \rightarrow 0}\left(1^{\operatorname{cosec}^2 t}+2^{\operatorname{cosec}^2 t}+\ldots \ldots . .+n^{\operatorname{cosec}^2 t}\right)^{\sin ^2 t} \\\\ & =\lim _{t \rightarrow 0} n\left(\left(\frac{1}{n}\right)^{\operatorname{cosec}^2 t}+\left(\frac{2}{n}\right)^{\operatorname{cosec}^2 t}+\ldots \ldots . .+1\right)^{\sin ^2 t} \\\\ & =\mathrm{n} \end{aligned}t→0lim(1cosec2t+2cosec2t+……..+ncosec2t)sin2t=t→0limn((n1)cosec2t+(n2)cosec2t+……..+1)sin2t=n