_ n→∞ npsin^2(n!) n+1 , 0 < p < 1 is equal to: - Vidyadip

MCQ

$\lim\limits_{\text{n}→∞} \frac{{\text{np}\text{sin}^2(\text{n}!)}}{\text{n}+1}$, 0 < p < 1 is equal to:

A
0
B
∞
C
1
D
None

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0

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