Discussion Forum : Limits Continuity And Differentiability
Question -


The value oflimx01cos3xxsin xcosx

Options:
A .   2/5
B .   3/5
C .   3/2
D .   3/4
Answer: Option C
:
C

limx01cos3xxsin xcosx=limx0(1cos x)(1+cos x+cos2x)xsin xcos x=limx02sin2(x2)2sin(x2)cos(x2).x×(1+cosx+cos2x)cos x=limx0sin(x2)2(x2)×1+cosx+cos2xcos(x2)cos x=12×3=32



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