bemath-lim-x-pi-2-1-sin-x-cos-x-

Question Number 111528 by bemath last updated on 04/Sep/20

\sqrt{bemath}

\lim_{x \to π / 2} {(1 - \sin x)}^{\cos x} ?

Answered by bobhans last updated on 04/Sep/20

let 1 - \sin x = w; where w \to 0 and \sin x = 1 - w

\cos x = \sqrt{1 - (1 - 2 w + w^{2})} = \sqrt{2 w - w^{2}}

L = \lim_{w \to 0} {(w)}^{\sqrt{2 w - w^{2}}}

\ln L = \lim_{w \to 0} \sqrt{2 w - w^{2}} \ln (w)

\ln L = \lim_{w \to 0} \frac{\ln (w)}{{(2 w - w^{2})}^{- \frac{1}{2}}}

\ln L = \lim_{w \to 0} \frac{\frac{1}{w}}{- \frac{1}{2} (2 - 2 w) {(2 w - w^{2})}^{- \frac{3}{2}}}

\ln L = \lim_{w \to 0} \frac{- {(2 w - w^{2})}^{\frac{3}{2}}}{w - w^{2}}

\ln L = \lim_{w \to 0} \frac{- \frac{3}{2} (2 - 2 w) \sqrt{2 w - w^{2}}}{1 - 2 w} = 0

∴ L = e^{0} = 1

Answered by Dwaipayan Shikari last updated on 04/Sep/20

\lim_{x \to \frac{π}{2}} {(1 - s i n x)}^{c o s x} = y

c o s x l o g (1 - s i n x) = l o g y

\frac{1}{2} (- 2 c o s x s i n x \frac{l o g (1 - s i n x)}{- s i n x}) = l o g y

\frac{1}{2} (- s i n 2 x) = l o g y

l o g y = 0

y = 1

Answered by 1549442205PVT last updated on 04/Sep/20

I = \lim_{x \to \frac{π}{2}} {(1 - \sin x)}^{\cos x}

lnI = \lim_{x \to \frac{π}{2}} [cosxln (1 - sinx)]

= \lim_{x \to \frac{π}{2}} \frac{\ln (1 - sinx)}{\frac{1}{cosx}} = \lim_{x \to \frac{π}{2}} \frac{\frac{1}{1 - sinx} \times (- cosx)}{\frac{sinx}{\cos^{2} x}}

(Since \lim_{x \to \frac{π}{2}} \frac{\ln (1 - sinx)}{\frac{1}{cosx}} = \frac{\infty}{\infty}) \Rightarrow using L^{'} Hopital

= \lim_{x \to \frac{π}{2}} \frac{- \cos^{3} x}{(1 - sinx) sinx} = \lim_{x \to π / 2} \frac{{3 \cos}^{2} xsinx}{(1 - sinx) cosx - cosxsinx}

= \lim_{x \to \frac{π}{2}} \frac{3 cosxsinx}{1 - 2 sinx} = \frac{0}{- 1} = 0 \Rightarrow I = e^{0} = 1