limx→0 11/sin2⁡x+21/sin2⁡x+……+n1/sin2⁡xsin2⁡x=

Given limit =limx→0 n⋅1n1sin2⁡x+2n1sin2⁡x+…+n−1n1sin2⁡x+1sin2⁡x=n(0+0+…+0+1)=n

limx→0 11/sin2⁡x+21/sin2⁡x+……+n1/sin2⁡xsin2⁡x=

Given limit =limx→0 n⋅1n1sin2⁡x+2n1sin2⁡x+…+n−1n1sin2⁡x+1sin2⁡x=n(0+0+…+0+1)=n

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$lim_{x \to 0} {[1^{1 / \sin^{2} x} + 2^{1 / \sin^{2} x} + \dots \dots + n^{1 / \sin^{2} x}]}^{\sin^{2} x} =$

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$\infty$

$\frac{n + 1}{2}$

answer is D.

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Detailed Solution

$\begin{array}{l} Given limit \\ = lim_{x \to 0} n \cdot {[{(\frac{1}{n})}^{\frac{1}{\sin^{2} x}} + {(\frac{2}{n})}^{\frac{1}{\sin^{2} x}} + \dots + {(\frac{n - 1}{n})}^{\frac{1}{\sin^{2} x}} + 1]}^{\sin^{2} x} \\ = n (0 + 0 + \dots + 0 + 1) = n \end{array}$