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Question Number 165868 by cortano1 last updated on 09/Feb/22

     lim_(x→(π/2))  ((cos x)/(sin x−((sin x+cos x))^(1/3) )) =?

limxπ2cosxsinxsinx+cosx3=?

Answered by qaz last updated on 10/Feb/22

lim_(x→(π/2)) ((cos x)/(sin x−((sin x+cos x))^(1/3) ))  =lim_(x→0) ((−sin x)/(cos x−((cos x−sin x))^(1/3) ))  =lim_(x→0) ((−x)/((1/3)ln(1−sin x)))  =−3lim_(x→0) (x/(sin x))  =−3

limxπ2cosxsinxsinx+cosx3=limx0sinxcosxcosxsinx3=limx0x13ln(1sinx)=3limx0xsinx=3

Answered by Rohit143Jo last updated on 10/Feb/22

Ans:  lim_(x→(π/2))   ((Cos x)/(Sin x − ((Sin x + Cos x))^(1/3) ))         = lim_(x→(π/2))   (((−Sin x))/(Cos x − ((Cos x − Sin x)/(3 (Sin x +Cos x)^(2/3) )))) [L′Hospital Rule]         = lim_(x→(π/2))   ((3.(−Sin x).(Sin x + Cos x)^(2/3) )/(3Cos x.(Sin x + Cos x)^(2/3)  − Cos x + Sin x))          =  ((3.(−Sin (π/2)).(Sin (π/2) + Cos (π/2))^(2/3) )/(3.Cos (π/2).(Sin (π/2) + Cos (π/2))^(2/3)  − Cos (π/2) + Sin (π/2)))         = (−3).

Ans:limxπ2CosxSinxSinx+Cosx3=limxπ2(Sinx)CosxCosxSinx3(Sinx+Cosx)23[LHospitalRule]=limxπ23.(Sinx).(Sinx+Cosx)233Cosx.(Sinx+Cosx)23Cosx+Sinx=3.(Sinπ2).(Sinπ2+Cosπ2)233.Cosπ2.(Sinπ2+Cosπ2)23Cosπ2+Sinπ2=(3).

Answered by greogoury55 last updated on 15/Feb/22

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