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Question Number 80000 by malwaan last updated on 30/Jan/20

prove that lim_(x→0) ((arcsin(x/(√(1−x^2 ))))/(ln(1−x))) = −1

$$\boldsymbol{{prove}}\:\boldsymbol{{that}} \\ $$$$\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\frac{\boldsymbol{{arcsin}}\frac{\boldsymbol{{x}}}{\sqrt{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} }}}{\boldsymbol{{ln}}\left(\mathrm{1}−\boldsymbol{{x}}\right)}\:=\:−\mathrm{1} \\ $$

Commented by jagoll last updated on 30/Jan/20

sin y = (x/(√(1−x^2 ))) ⇒(x^2 /(1−x^2 )) = sin^2 y ((1−x^2 −1)/(1−x^2 ))=sin^2 y ⇒1−(1/(1−x^2 ))=sin^2 y cos^2 y=(1/(1−x^2 )) ⇒1−x^2 =sec^2 y 1−x=((sec^2 y)/(1+x))

$$\mathrm{sin}\:{y}\:=\:\frac{{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\Rightarrow\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }\:=\:\mathrm{sin}\:^{\mathrm{2}} {y} \\ $$$$\frac{\mathrm{1}−{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }=\mathrm{sin}\:^{\mathrm{2}} {y}\:\Rightarrow\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }=\mathrm{sin}\:^{\mathrm{2}} {y} \\ $$$$\mathrm{cos}\:^{\mathrm{2}} {y}=\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\:\Rightarrow\mathrm{1}−{x}^{\mathrm{2}} =\mathrm{sec}\:^{\mathrm{2}} {y} \\ $$$$\mathrm{1}−{x}=\frac{\mathrm{sec}\:^{\mathrm{2}} {y}}{\mathrm{1}+{x}} \\ $$$$ \\ $$

Commented by abdomathmax last updated on 30/Jan/20

let f(x)=((arcsin((x/(√(1−x^2 )))))/(ln(1−x))) changement x=sint give f(x)=((arcsin(tant))/(ln(1−sint)))=g(t) x→0 ⇒t→0 ⇒ g(t)∼((arcsint)/(−t)) =−((arcsint)/t) and lim_(t→0) ((arsin(t))/t) = arcsin^′ (0) we have arcsin^′ (t) = (1/(√(1−t^2 ))) ⇒arcsin^′ (0) =1 ⇒ lim_(t→0) g(t)=−1 =lim_(x→0) f(x)

$${let}\:{f}\left({x}\right)=\frac{{arcsin}\left(\frac{{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right)}{{ln}\left(\mathrm{1}−{x}\right)}\:\:{changement}\:{x}={sint}\: \\ $$$${give}\:{f}\left({x}\right)=\frac{{arcsin}\left({tant}\right)}{{ln}\left(\mathrm{1}−{sint}\right)}={g}\left({t}\right) \\ $$$${x}\rightarrow\mathrm{0}\:\Rightarrow{t}\rightarrow\mathrm{0}\:\:\:\Rightarrow\:{g}\left({t}\right)\sim\frac{{arcsint}}{−{t}}\:=−\frac{{arcsint}}{{t}} \\ $$$${and}\:{lim}_{{t}\rightarrow\mathrm{0}} \:\frac{{arsin}\left({t}\right)}{{t}}\:=\:{arcsin}^{'} \left(\mathrm{0}\right)\:\:{we}\:{have} \\ $$$${arcsin}^{'} \left({t}\right)\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:\Rightarrow{arcsin}^{'} \left(\mathrm{0}\right)\:=\mathrm{1}\:\Rightarrow \\ $$$${lim}_{{t}\rightarrow\mathrm{0}} \:{g}\left({t}\right)=−\mathrm{1}\:={lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right) \\ $$

Answered by Kamel Kamel last updated on 30/Jan/20

Ω=lim_(x→0) ((arcsin((x/(√(1−x^2 )))))/(Ln(1−x))) Put: x=sin(t),x→0⇒t→0 Ω=lim_(x→0) ((arcsin((x/(√(1−x^2 )))))/(Ln(1−x)))=lim_(t→0) ((arcsin(tan(t)))/(Ln(1−sin(t))))=−lim((tan(t))/(sin(t)))=−1

$$\Omega=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{arcsin}\left(\frac{\mathrm{x}}{\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\right)}{\mathrm{Ln}\left(\mathrm{1}−\mathrm{x}\right)} \\ $$$$\mathrm{Put}:\:\mathrm{x}=\mathrm{sin}\left(\mathrm{t}\right),\mathrm{x}\rightarrow\mathrm{0}\Rightarrow\mathrm{t}\rightarrow\mathrm{0} \\ $$$$\Omega=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{arcsin}\left(\frac{\mathrm{x}}{\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\right)}{\mathrm{Ln}\left(\mathrm{1}−\mathrm{x}\right)}=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{arcsin}\left(\mathrm{tan}\left(\mathrm{t}\right)\right)}{\mathrm{Ln}\left(\mathrm{1}−\mathrm{sin}\left(\mathrm{t}\right)\right)}=−\mathrm{lim}\frac{\mathrm{tan}\left(\mathrm{t}\right)}{\mathrm{sin}\left(\mathrm{t}\right)}=−\mathrm{1} \\ $$

Commented by malwaan last updated on 31/Jan/20

thank you so much

$${thank}\:{you}\:{so}\:{much} \\ $$

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