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Ltx→01−cos x cos 2x cos 3xsin22x=

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a

32

b

52

c

74

d

92

answer is C.

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Detailed Solution

limx→0 1-cosxcos2xcos3x(sin22x/4x2)4x2=limx→0 1-cosxcos2xcos3x4x2 =0/0 use L hospital rule limx→0 0+sinxcos2xcos3x+2sin2xcosxcos3x+3sin3xcosxcos2x8x=1+4+9/8=14/8

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