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JEE Main 2018
Limits, Continuity & Differentiability
Limits, Continuity and Differentiability
Medium

Question

limt0(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}} is equal to

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Solution

limt0(1cosec2t+2cosec2t+..+ncosec2t)sin2t=limt0n((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}

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