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1-cosxcos2xcos3x = 1-(1/2)cos2x(cos4x+cos2x)
= 1-(1/2)cos2xcos4x-(1/2)(cos2x)^2
=1-(1/4)(cos6x+cos2x)-(1/4)(1+cos4x)
= 3/4 - (1/4)(cos6x+cos4x+cos2x)
lim<x→0>(1-cosxcos2xcos3x)/(ax^n)
= lim<x→0>[3/4 - (1/4)(cos6x+cos4x+cos2x)]/(ax^n) (0/0)
= lim<x→0>(1/2)(3sin6x+2sin4x+sin2x)]/[anx^(n-1)] (0/0)
= lim<x→0>(9cos6x+4cos4x+cos2x)]/[an(n-1)x^(n-2)] = 1
n-2 = 0, an(n-1) = 14, n = 2, a = 7
= 1-(1/2)cos2xcos4x-(1/2)(cos2x)^2
=1-(1/4)(cos6x+cos2x)-(1/4)(1+cos4x)
= 3/4 - (1/4)(cos6x+cos4x+cos2x)
lim<x→0>(1-cosxcos2xcos3x)/(ax^n)
= lim<x→0>[3/4 - (1/4)(cos6x+cos4x+cos2x)]/(ax^n) (0/0)
= lim<x→0>(1/2)(3sin6x+2sin4x+sin2x)]/[anx^(n-1)] (0/0)
= lim<x→0>(9cos6x+4cos4x+cos2x)]/[an(n-1)x^(n-2)] = 1
n-2 = 0, an(n-1) = 14, n = 2, a = 7
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