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JEE MATHEMATICS MCQ: LIMITS

1 Easy
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Analyzing +4 | -1
The value of lim⁡x→0{11sin⁡2x+21sin⁡2x+31sin⁡2x+….+n1sin⁡2x}sin⁡2x,\lim \limits_{{x \rightarrow 0}}\left\{1^{\frac{1}{\sin ^2 x}+} 2^{\frac{1}{\sin ^2 x}}+3^{\frac{1}{\sin ^2 x}}+\ldots .+n^{\frac{1}{\sin ^2 x}}\right\}^{\sin ^2 x}, is
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Correct option: C
nn
Option C (n) is correct because, as x approaches 0, the expression simplifies to n.

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