The value oflimx→01−cos3xxsin xcosx
2/5
3/5
3/2
3/4
limx→01−cos3xxsin xcosx=limx→0(1−cos x)(1+cos x+cos2x)xsin xcos x=limx→02sin2(x2)2sin(x2)cos(x2).x×(1+cosx+cos2x)cos x=limx→0sin(x2)2(x2)×1+cosx+cos2xcos(x2)cos x=12×3=32
limx→0 1−cos3xsin3x.sin5x =