JEE MATHEMATICS MCQ: LIMITS

1 Easy

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Analyzing +4 | -1

\lim \limits_{{n \rightarrow \infty}} \frac{n^p \sin ^2(n !)}{n+1}, 0 < p < 1

, is equal to

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Correct option: A

Option A (0) is correct because the expression can be simplified to an oscillating number divided by infinity, which approaches 0.

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