Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 66318 by mathmax by abdo last updated on 12/Aug/19

find lim_(x→0) (((1+x)/(1−x)))^(1/(sinx))

$${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)^{\frac{\mathrm{1}}{{sinx}}} \\ $$

Commented by mathmax by abdo last updated on 24/Aug/19

let f(x)=(((1+x)/(1−x)))^(1/(sinx)) ⇒f(x)=e^((1/(sinx))ln(((1+x)/(1−x)))) but ln(((1+x)/(1−x))) =ln(1+x)−ln(1−x) ∼x−(−x) =2x and sinx∼x ⇒ (1/(sinx))ln(((1+x)/(1−x))) ∼(1/x)(2x) =2 ⇒ lim_(x→0) f(x)=e^2

$${let}\:{f}\left({x}\right)=\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)^{\frac{\mathrm{1}}{{sinx}}} \:\Rightarrow{f}\left({x}\right)={e}^{\frac{\mathrm{1}}{{sinx}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)} \:\:{but} \\ $$$${ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)\:={ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{1}−{x}\right)\:\sim{x}−\left(−{x}\right)\:=\mathrm{2}{x}\:{and}\:{sinx}\sim{x}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{{sinx}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)\:\sim\frac{\mathrm{1}}{{x}}\left(\mathrm{2}{x}\right)\:=\mathrm{2}\:\Rightarrow\:{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)={e}^{\mathrm{2}} \\ $$

Commented by kaivan.ahmadi last updated on 12/Aug/19

=e^(lim_(x→0) (((1+x)/(1−x))−1)×(1/x)) =e^(lim_(x→0) (((2x)/(1−x)))×(1/x)) =e^2

$$={e}^{{lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}−\mathrm{1}\right)×\frac{\mathrm{1}}{{x}}} ={e}^{{lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{\mathrm{2}{x}}{\mathrm{1}−{x}}\right)×\frac{\mathrm{1}}{{x}}} ={e}^{\mathrm{2}} \\ $$$$ \\ $$

Commented by Mikael last updated on 12/Aug/19

lim_(x→0) [1+(((1+x)/(1−x))−1)]^(1/(sinx)) = lim_(x→0) [(1+((2x)/(1−x)))^((1−x)/(2x)) ]^(((2x)/(1−x)).(1/(sinx))) lim_(x→0) [(1+((2x)/(1−x)))^((1−x)/(2x)) ]=e ,then lim_(x→0) ((2x)/(1−x)).(1/(sinx)) =lim_(x→0) (x/(sinx)).(2/(1−x))= 1×2=2 = e^2

$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left[\mathrm{1}+\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}−\mathrm{1}\right)\right]^{\frac{\mathrm{1}}{{sinx}}} =\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\left[\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{1}−{x}}\right)^{\frac{\mathrm{1}−{x}}{\mathrm{2}{x}}} \right]^{\frac{\mathrm{2}{x}}{\mathrm{1}−{x}}.\frac{\mathrm{1}}{{sinx}}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\left[\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{1}−{x}}\right)^{\frac{\mathrm{1}−{x}}{\mathrm{2}{x}}} \right]={e}\:,{then}\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\mathrm{2}{x}}{\mathrm{1}−{x}}.\frac{\mathrm{1}}{{sinx}}\:=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{{x}}{{sinx}}.\frac{\mathrm{2}}{\mathrm{1}−{x}}=\:\mathrm{1}×\mathrm{2}=\mathrm{2} \\ $$$$=\:{e}^{\mathrm{2}} \: \\ $$

Commented by mathmax by abdo last updated on 12/Aug/19

thanks sir.

$${thanks}\:{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com