Solve Q 6 plz dont give link
6 . lim X 0   ln ( sin 3 x ) ln ( sin x )   i s   e q u a l   t o A )   0 B )   1 C )   2 D )   N o n   e x i s t e n t

Dear student
limx0lnsin3xlnsinxApply L'Hopital's Rule=limx03cos3xsin3xcosxsinx=limx03cot3xcotxAgain, apply L'Hopital's Rule=limx0-9cosec23x-cosec2x=limx09sin23x1sin2x=limx09sin2xsin23xAgain, apply L'Hopital's Rule=limx09×2sinx×cosx2×sin3x×cos3x×3=limx09×sin2xsin6x×3=limx03sin2xsin6xAgain, apply L'Hopital's Rule=limx06cos2x6cos6x=limx0cos2xcos6xPutting x=0 , we get=cos2×0cos6×0=cos0cos0=11=1Note:sin2A=2sinAcosA
Regards

  • 0
What are you looking for?