Menu Close

calculate-lim-x-1-1-1-x-arcsinx-pi-2-

Tinku Tara 0 Comments
Pin




Question Number 33594 by abdo imad last updated on 19/Apr/18
calculate lim_(x→1^− ) (1/((1−x)^α ))(arcsinx −(π/2)) .
$${calculate}\:\:{lim}_{{x}\rightarrow\mathrm{1}^{−} } \:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\alpha} }\left({arcsinx}\:−\frac{\pi}{\mathrm{2}}\right)\:. \\ $$
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....
$${let}\:{use}\:{the}\:{ch}.\:\mathrm{1}−{x}\:={t}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{1}^{−} } \:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\alpha} }\left(\:{arcsinx}\:−\frac{\pi}{\mathrm{2}}\right) \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}^{+} } \:\:\:\frac{\mathrm{1}}{{t}^{\alpha} }\left({arcsin}\left(\mathrm{1}−{t}\right)\:−\frac{\pi}{\mathrm{2}}\right) \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}^{+} } \:\:\:\frac{{arcsin}\left(\mathrm{1}−{t}\right)−\frac{\pi}{\mathrm{2}}}{{t}^{\alpha} }\:\:{let}\:{use}\:{hospital}\:{theorem} \\ $$$${u}\left({t}\right)\:={arcsin}\left(\mathrm{1}−{t}\right)−\frac{\pi}{\mathrm{2}}\:{and}\:{v}\left({t}\right)=\:{t}^{\alpha} \\ $$$${u}^{'} \left({t}\right)\:=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{1}−\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }}\:\:{and}\:{v}^{'} \left({t}\right)\:=\alpha\:{t}^{\alpha−\mathrm{1}} \Rightarrow \\ $$$${u}^{'} \left({t}\right)\:=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} \:+\mathrm{2}{t}\:−\mathrm{1}}}\:=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{t}\:−{t}^{\mathrm{2}} }}\:{if}\:\alpha>\mathrm{1}\:{lim}\:\alpha\:{t}^{\alpha−\mathrm{1}} \:=\mathrm{0}^{+} \\ $$$${lim}_{{t}\rightarrow\mathrm{0}^{+} } \:\:\:{u}^{'} \left({t}\right)\:=−\infty\:\Rightarrow{lim}\:_{{t}\rightarrow{o}^{+} } \:\:\:\frac{{arcsin}\left(\mathrm{1}−{t}\right)−\frac{\pi}{\mathrm{2}}}{{t}^{\alpha} }\:=−\infty \\ $$$${if}\:\mathrm{0}<\alpha<\mathrm{1}\:{we}\:{do}\:{the}\:{ch}.\alpha\:=\frac{\mathrm{1}}{\lambda}\:{with}\:\lambda>\mathrm{1}\Rightarrow \\ $$$${lim}_{{t}\rightarrow\mathrm{0}^{+} } \:\:\:\:\frac{{arcsin}\left(\mathrm{1}−{t}\right)−\frac{\pi}{\mathrm{2}}}{{t}^{\alpha} }\:={lim}_{{t}\rightarrow\mathrm{0}^{+} } \:\:\frac{{arcsin}\left(\mathrm{1}−{t}\right)−\frac{\pi}{\mathrm{2}}}{{t}^{\frac{\mathrm{1}}{\lambda}} } \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}^{+} } \:\:\:\:\:\:\frac{−\mathrm{1}}{\frac{\mathrm{1}}{\lambda}{t}^{\frac{\mathrm{1}}{\lambda}−\mathrm{1}} \sqrt{\mathrm{2}{t}−{t}^{\mathrm{2}} }}\:={lim}_{{t}\rightarrow\mathrm{0}^{+} } \:\:\:\frac{−\lambda}{\:\sqrt{\mathrm{2}{t}−{t}^{\mathrm{2}} }}\:{t}^{\mathrm{1}−\frac{\mathrm{1}}{\lambda}} \:{for} \\ $$$${that}\:{we}\:{must}\:{calculate}\:{u}^{''} \left({t}\right)\:{and}\:{v}^{''} \left({t}\right)…{be}\:{continued}…. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *