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Question Number 165868 by cortano1 last updated on 09/Feb/22
limx→π2cosxsinx−sinx+cosx3=?
Answered by qaz last updated on 10/Feb/22
limx→π2cosxsinx−sinx+cosx3=limx→0−sinxcosx−cosx−sinx3=limx→0−x13ln(1−sinx)=−3limx→0xsinx=−3
Answered by Rohit143Jo last updated on 10/Feb/22
Ans:limx→π2CosxSinx−Sinx+Cosx3=limx→π2(−Sinx)Cosx−Cosx−Sinx3(Sinx+Cosx)23[L′HospitalRule]=limx→π23.(−Sinx).(Sinx+Cosx)233Cosx.(Sinx+Cosx)23−Cosx+Sinx=3.(−Sinπ2).(Sinπ2+Cosπ2)233.Cosπ2.(Sinπ2+Cosπ2)23−Cosπ2+Sinπ2=(−3).
Answered by greogoury55 last updated on 15/Feb/22
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