Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com