lim _{t} {rightarrow 0}left(1^{frac{1}{sin ^2 | vedclass

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(B) माना $L = \lim _{t}$ ${\rightarrow 0}\left(1^{\operatorname{cosec}^2 t}+2^{\operatorname{cosec}^2 t}+\ldots +n^{\operatorname{cosec}^2 t}\right)^{

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$n$

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(B) माना $L = \lim _{t}$ ${\rightarrow 0}\left(1^{\operatorname{cosec}^2 t}+2^{\operatorname{cosec}^2 t}+\ldots +n^{\operatorname{cosec}^2 t}\right)^{\sin ^2 t}$ है।कोष्ठक से सबसे बड़ा पद $n^{\operatorname{cosec}^2 t}$ बाहर लेने पर:$L = \lim _{t}$ ${\rightarrow 0} \left[ n^{\operatorname{cosec}^2 t} \left( \left(\frac{1}{n}\right)^{\operatorname{cosec}^2 t} + \left(\frac{2}{n}\right)^{\operatorname{cosec}^2 t} + \ldots + 1 \right) \right]^{\sin ^2 t}$.$L = \lim _{t}$ ${\rightarrow 0} \left( n^{\operatorname{cosec}^2 t} \right)^{\sin ^2 t} \cdot \left( \left(\frac{1}{n}\right)^{\operatorname{cosec}^2 t} + \left(\frac{2}{n}\right)^{\operatorname{cosec}^2 t} + \ldots + 1 \right)^{\sin ^2 t}$.चूंकि $\operatorname{cosec}^2 t \cdot \sin ^2 t = 1$,पहला भाग $n^1 = n$ हो जाता है।दूसरे भाग के लिए,जैसे $t \rightarrow 0$,$\operatorname{cosec}^2 t \rightarrow \infty$ होता है। इसलिए,$k अतः,व्यंजक $n \cdot (0 + 0 + \ldots + 1)^0 = n \cdot 1^0 = n$ हो जाता है।

$\lim _{t}$ ${\rightarrow 0}\left(1^{\frac{1}{\sin ^2 t}}+2^{\frac{1}{\sin ^2 t}}+\ldots +n^{\frac{1}{\sin ^2 t}}\right)^{\sin ^2 t}$ का मान $.......$ है।

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