The value of the limit limx→0 11/sin2⁡x+21/sin2⁡x+…+n1/sin2⁡xsin2⁡x is

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∞

n(n+1)2

answer is D.

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Detailed Solution

limx→0 11/sin2⁡x+21/sin2⁡x+…+n1/sin2⁡xsin2⁡xput 1sin2⁡x=t≥1∴limt→∞ 1t+2t+…+nt1/t =limt→∞ nt1/t1nt+2nt+…+11/t =nlimt→∞ 1nt+2nt+…+11/t=n[0+0+…+1]0=n

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