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

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

Vedclass · Accepted Answer

$n$

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(B) Let $L = \lim _{t}$ ${\rightarrow 0}\left(1^{\operatorname{cosec}^2 t}+2^{\operatorname{cosec}^2 t}+\ldots +n^{\operatorname{cosec}^2 t}\right)^{\sin ^2 t}$.We can factor out the largest term $n^{\operatorname{cosec}^2 t}$ from the bracket:$L = \lim _{t}$ ${\rightarrow 0} \left[ n^{\operatorname{cosec}^2 t} \left( \left(\frac{1}{n}\right)^{\operatorname{cosec}^2 t} + \left(\frac{2}{n}\right)^{\operatorname{cosec}^2 t} + \ldots + 1 \right) \right]^{\sin ^2 t}$.$L = \lim _{t}$ ${\rightarrow 0} \left( n^{\operatorname{cosec}^2 t} \right)^{\sin ^2 t} \cdot \left( \left(\frac{1}{n}\right)^{\operatorname{cosec}^2 t} + \left(\frac{2}{n}\right)^{\operatorname{cosec}^2 t} + \ldots + 1 \right)^{\sin ^2 t}$.Since $\operatorname{cosec}^2 t \cdot \sin ^2 t = 1$,the first part becomes $n^1 = n$.For the second part,as $t \rightarrow 0$,$\operatorname{cosec}^2 t \rightarrow \infty$. Thus,for $k So,the expression becomes $n \cdot (0 + 0 + \ldots + 1)^0 = n \cdot 1^0 = n \cdot 1 = n$.

$\lim _{t}$ ${\rightarrow 0}\left(1^{\frac{1}{\sin ^2 t}}+2^{\frac{1}{\sin ^2 t}}+\ldots +n^{\frac{1}{\sin ^2 t}}\right)^{\sin ^2 t}$ is equal to $.......$

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