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

Question Number 33594 by abdo imad last updated on 19/Apr/18

c a l c u l a t e {l i m}_{x \to 1^{-}} \frac{1}{{(1 - x)}^{α}} (a r c s i n x - \frac{π}{2}) .

Commented by abdo imad last updated on 24/Apr/18

l e t u s e t h e c h . 1 - x = t \Rightarrow {l i m}_{x \to 1^{-}} \frac{1}{{(1 - x)}^{α}} (a r c s i n x - \frac{π}{2})

= {l i m}_{t \to 0^{+}} \frac{1}{t^{α}} (a r c s i n (1 - t) - \frac{π}{2})

= {l i m}_{t \to 0^{+}} \frac{a r c s i n (1 - t) - \frac{π}{2}}{t^{α}} l e t u s e h o s p i t a l t h e o r e m

u (t) = a r c s i n (1 - t) - \frac{π}{2} a n d v (t) = t^{α}

u^{^{'}} (t) = \frac{- 1}{\sqrt{1 - {(1 - t)}^{2}}} a n d v^{^{'}} (t) = α t^{α - 1} \Rightarrow

u^{^{'}} (t) = \frac{- 1}{\sqrt{1 - t^{2} + 2 t - 1}} = - \frac{1}{\sqrt{2 t - t^{2}}} i f α > 1 l i m α t^{α - 1} = 0^{+}

{l i m}_{t \to 0^{+}} u^{^{'}} (t) = - \infty \Rightarrow l i m_{t \to o^{+}} \frac{a r c s i n (1 - t) - \frac{π}{2}}{t^{α}} = - \infty

i f 0 < α < 1 w e d o t h e c h . α = \frac{1}{λ} w i t h λ > 1 \Rightarrow

{l i m}_{t \to 0^{+}} \frac{a r c s i n (1 - t) - \frac{π}{2}}{t^{α}} = {l i m}_{t \to 0^{+}} \frac{a r c s i n (1 - t) - \frac{π}{2}}{t^{\frac{1}{λ}}}

= {l i m}_{t \to 0^{+}} \frac{- 1}{\frac{1}{λ} t^{\frac{1}{λ} - 1} \sqrt{2 t - t^{2}}} = {l i m}_{t \to 0^{+}} \frac{- λ}{\sqrt{2 t - t^{2}}} t^{1 - \frac{1}{λ}} f o r

t h a t w e m u s t c a l c u l a t e u^{^{″}} (t) a n d v^{^{″}} (t) \dots b e c o n t i n u e d \dots .