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

$$\sqrt{\mathrm{bemath}}$$$$\underset{x\to \pi /2}{\lim }{(1-\sin \mathrm{x})}^{\cos \mathrm{x}}?$$

Answered by bobhans last updated on 04/Sep/20

$$\mathrm{let}1-\sin \mathrm{x}=\mathrm{w};\mathrm{where}\mathrm{w}\to 0\mathrm{and}\sin \mathrm{x}=1-\mathrm{w}$$$$\cos \mathrm{x}=\sqrt{1-(1-2\mathrm{w}+{\mathrm{w}}^2)}=\sqrt{2\mathrm{w}-{\mathrm{w}}^2}$$$$\mathrm{L}=\underset{\mathrm{w}\to 0}{\lim }{\left(\mathrm{w}\right)}^{\sqrt{2\mathrm{w}-{\mathrm{w}}^2}}$$$$\ln \mathrm{L}=\underset{\mathrm{w}\to 0}{\lim }\sqrt{2\mathrm{w}-{\mathrm{w}}^2}\ln \left(\mathrm{w}\right)$$$$\ln \mathrm{L}=\underset{\mathrm{w}\to 0}{\lim }\frac{\ln \left(\mathrm{w}\right)}{{(2\mathrm{w}-{\mathrm{w}}^2)}^{-\frac{1}{2}}}$$$$\ln \mathrm{L}=\underset{\mathrm{w}\to 0}{\lim }\frac{\frac{1}{\mathrm{w}}}{-\frac{1}{2}(2-2\mathrm{w}){(2\mathrm{w}-{\mathrm{w}}^2)}^{-\frac{3}{2}}}$$$$\ln \mathrm{L}=\underset{\mathrm{w}\to 0}{\lim }\frac{-{(2\mathrm{w}-{\mathrm{w}}^2)}^{\frac{3}{2}}}{\mathrm{w}-{\mathrm{w}}^2}$$$$\ln \mathrm{L}=\underset{\mathrm{w}\to 0}{\lim }\frac{-\frac{3}{2}(2-2\mathrm{w})\sqrt{2\mathrm{w}-{\mathrm{w}}^2}}{1-2\mathrm{w}}=0$$$$\therefore \mathrm{L}={\mathrm{e}}^0=1$$

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

$$\underset{x\to \frac{\pi }{2}}{\lim }{(1-sinx)}^{cosx}=y$$$$cosxlog(1-sinx)=logy$$$$\frac{1}{2}(-2cosxsinx\frac{log(1-sinx)}{-sinx})=logy$$$$\frac{1}{2}(-sin2x)=logy$$$$logy=0$$$$y=1$$

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

$$\mathrm{I}=\underset{\mathrm{x}\to \frac{\pi }{2}}{\lim }{(1-\sin \mathrm{x})}^{\cos \mathrm{x}}$$$$\mathrm{lnI}=\underset{\mathrm{x}\to \frac{\pi }{2}}{\lim }\left[\mathrm{cosxln}(1-\mathrm{sinx})\right]$$$$=\underset{\mathrm{x}\to \frac{\pi }{2}}{\lim }\frac{\ln (1-\mathrm{sinx})}{\frac{1}{\mathrm{cosx}}}=\underset{\mathrm{x}\to \frac{\pi }{2}}{\lim }\frac{\frac{1}{1-\mathrm{sinx}}\times (-\mathrm{cosx})}{\frac{\mathrm{sinx}}{{\cos }^2\mathrm{x}}}$$$$(\mathrm{Since}\underset{\mathrm{x}\to \frac{\pi }{2}}{\lim }\frac{\ln (1-\mathrm{sinx})}{\frac{1}{\mathrm{cosx}}}=\frac{\mathrm{\infty }}{\mathrm{\infty }})\Rightarrow \mathrm{using}{\mathrm{L}}^{\prime }\mathrm{Hopital}$$$$=\underset{\mathrm{x}\to \frac{\pi }{2}}{\lim }\frac{-{\cos }^3\mathrm{x}}{(1-\mathrm{sinx})\mathrm{sinx}}=\underset{\mathrm{x}\to \pi /2}{\lim }\frac{{3\cos }^2\mathrm{xsinx}}{(1-\mathrm{sinx})\mathrm{cosx}-\mathrm{cosxsinx}}$$$$=\underset{\mathrm{x}\to \frac{\pi }{2}}{\mathbf{lim}}\frac{3\mathbf{cosxsinx}}{1-2\mathbf{sinx}}=\frac{0}{-1}=0\Rightarrow \mathbf{I}={\mathbf{e}}^0=1$$

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