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

(√(bemath)) lim_(x→π/2) (1−sin x)^(cos x) ?

$$\:\:\:\sqrt{\mathrm{bemath}} \\ $$$$\:\:\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\left(\mathrm{1}−\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{cos}\:\mathrm{x}} \:? \\ $$

Answered by bobhans last updated on 04/Sep/20

let 1−sin x = w ; where w→0 and sin x=1−w cos x = (√(1−(1−2w+w^2 ))) = (√(2w−w^2 )) L= lim_(w→0) (w)^(√(2w−w^2 )) ln L = lim_(w→0) (√(2w−w^2 )) ln (w) ln L = lim_(w→0) ((ln (w))/((2w−w^2 )^(−(1/2)) )) ln L= lim_(w→0) ((1/w)/(−(1/2)(2−2w)(2w−w^2 )^(−(3/2)) )) ln L = lim_(w→0) ((−(2w−w^2 )^(3/2) )/(w−w^2 )) ln L = lim_(w→0) ((−(3/2)(2−2w)(√(2w−w^2 )))/(1−2w)) = 0 ∴ L = e^0 = 1

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

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

lim_(x→(π/2)) (1−sinx)^(cosx) =y cosxlog(1−sinx)=logy (1/2)(−2cosxsinx((log(1−sinx))/(−sinx)))=logy (1/2)(−sin2x)=logy logy=0 y=1

$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\left(\mathrm{1}−{sinx}\right)^{{cosx}} ={y} \\ $$$${cosxlog}\left(\mathrm{1}−{sinx}\right)={logy} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{2}{cosxsinx}\frac{{log}\left(\mathrm{1}−{sinx}\right)}{−{sinx}}\right)={logy} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(−{sin}\mathrm{2}{x}\right)={logy} \\ $$$${logy}=\mathrm{0} \\ $$$${y}=\mathrm{1} \\ $$

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

I= lim_(x→(π/2)) (1−sin x)^(cos x) lnI=lim_(x→(π/2)) [cosxln(1−sinx)] =lim_(x→(π/2)) ((ln(1−sinx))/(1/(cosx)))=lim_(x→(π/2)) (((1/(1−sinx))×(−cosx))/((sinx)/(cos^2 x))) (Since lim_(x→(π/2)) ((ln(1−sinx))/(1/(cosx)))=(∞/∞))⇒using L′Hopital =lim_(x→(π/2)) ((−cos^3 x)/((1−sinx)sinx))=lim_(x→π/2) ((3cos^2 xsinx)/((1−sinx)cosx−cosxsinx)) =lim_(x→(π/2)) ((3cosxsinx)/(1−2sinx))=(0/(−1))=0⇒I=e^0 =1

$$\mathrm{I}=\:\:\underset{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\left(\mathrm{1}−\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{cos}\:\mathrm{x}} \: \\ $$$$\mathrm{lnI}=\underset{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\left[\mathrm{cosxln}\left(\mathrm{1}−\mathrm{sinx}\right)\right] \\ $$$$=\underset{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{sinx}\right)}{\frac{\mathrm{1}}{\mathrm{cosx}}}=\underset{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sinx}}×\left(−\mathrm{cosx}\right)}{\frac{\mathrm{sinx}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}} \\ $$$$\left(\mathrm{Since}\:\:\underset{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{sinx}\right)}{\frac{\mathrm{1}}{\mathrm{cosx}}}=\frac{\infty}{\infty}\right)\Rightarrow\mathrm{using}\:\mathrm{L}'\mathrm{Hopital} \\ $$$$=\underset{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\frac{−\mathrm{cos}^{\mathrm{3}} \mathrm{x}}{\left(\mathrm{1}−\mathrm{sinx}\right)\mathrm{sinx}}=\underset{\mathrm{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{3cos}^{\mathrm{2}} \mathrm{xsinx}}{\left(\mathrm{1}−\mathrm{sinx}\right)\mathrm{cosx}−\mathrm{cosxsinx}} \\ $$$$=\underset{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\boldsymbol{\mathrm{lim}}}\frac{\mathrm{3}\boldsymbol{\mathrm{cosxsinx}}}{\mathrm{1}−\mathrm{2}\boldsymbol{\mathrm{sinx}}}=\frac{\mathrm{0}}{−\mathrm{1}}=\mathrm{0}\Rightarrow\boldsymbol{\mathrm{I}}=\boldsymbol{\mathrm{e}}^{\mathrm{0}} =\mathrm{1} \\ $$

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