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Question Number 33594 by abdo imad last updated on 19/Apr/18
calculate lim_(x→1^− ) (1/((1−x)^α ))(arcsinx −(π/2)) .
calculatelimx→1−1(1−x)α(arcsinx−π2).
Commented by abdo imad last updated on 24/Apr/18
let use the ch. 1−x =t ⇒lim_(x→1^− ) (1/((1−x)^α ))( arcsinx −(π/2)) =lim_(t→0^+ ) (1/t^α )(arcsin(1−t) −(π/2)) =lim_(t→0^+ ) ((arcsin(1−t)−(π/2))/t^α ) let use hospital theorem u(t) =arcsin(1−t)−(π/2) and v(t)= t^α u^′ (t) =((−1)/( (√(1−(1−t)^2 )))) and v^′ (t) =α t^(α−1) ⇒ u^′ (t) =((−1)/( (√(1−t^2 +2t −1)))) =−(1/( (√(2t −t^2 )))) if α>1 lim α t^(α−1) =0^+ lim_(t→0^+ ) u^′ (t) =−∞ ⇒lim _(t→o^+ ) ((arcsin(1−t)−(π/2))/t^α ) =−∞ if 0<α<1 we do the ch.α =(1/λ) with λ>1⇒ lim_(t→0^+ ) ((arcsin(1−t)−(π/2))/t^α ) =lim_(t→0^+ ) ((arcsin(1−t)−(π/2))/t^(1/λ) ) =lim_(t→0^+ ) ((−1)/((1/λ)t^((1/λ)−1) (√(2t−t^2 )))) =lim_(t→0^+ ) ((−λ)/( (√(2t−t^2 )))) t^(1−(1/λ)) for that we must calculate u^(′′) (t) and v^(′′) (t)...be continued....
letusethech.1−x=t⇒limx→1−1(1−x)α(arcsinx−π2)=limt→0+1tα(arcsin(1−t)−π2)=limt→0+arcsin(1−t)−π2tαletusehospitaltheoremu(t)=arcsin(1−t)−π2andv(t)=tαu′(t)=−11−(1−t)2andv′(t)=αtα−1⇒u′(t)=−11−t2+2t−1=−12t−t2ifα>1limαtα−1=0+limt→0+u′(t)=−∞⇒limt→o+arcsin(1−t)−π2tα=−∞if0<α<1wedothech.α=1λwithλ>1⇒limt→0+arcsin(1−t)−π2tα=limt→0+arcsin(1−t)−π2t1λ=limt→0+−11λt1λ−12t−t2=limt→0+−λ2t−t2t1−1λforthatwemustcalculateu″(t)andv″(t)…becontinued….

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