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Question Number 58438 by salahahmed last updated on 23/Apr/19

lim_(x→∞) (((x−1)/x))^x

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{x}−\mathrm{1}}{{x}}\right)^{{x}} \\ $$

Commented by mr W last updated on 23/Apr/19

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

$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\frac{{x}}{{x}−\mathrm{1}}}\right)^{{x}} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\left(\frac{{x}}{{x}−\mathrm{1}}\right)\left(\frac{{x}}{{x}−\mathrm{1}}\right)^{{x}−\mathrm{1}} } \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{{x}}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{x}−\mathrm{1}}\right)^{{x}−\mathrm{1}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}×{e}} \\ $$$$=\frac{\mathrm{1}}{{e}} \\ $$

Commented by salahahmed last updated on 23/Apr/19

Why does (((1/x)/(x−1)))^x =((((x/(x−1))))/(((x/(x−1)))^(x−1) )) and ((x/(x−1)))^(x−1) =(1+(1/(x−1)))^(x−1)

$$\mathrm{Why}\:\mathrm{does}\:\left(\frac{\frac{\mathrm{1}}{{x}}}{{x}−\mathrm{1}}\right)^{{x}} =\frac{\left(\frac{{x}}{{x}−\mathrm{1}}\right)}{\left(\frac{{x}}{{x}−\mathrm{1}}\right)^{{x}−\mathrm{1}} }\:\mathrm{and}\:\left(\frac{{x}}{{x}−\mathrm{1}}\right)^{{x}−\mathrm{1}} =\left(\mathrm{1}+\frac{\mathrm{1}}{{x}−\mathrm{1}}\right)^{{x}−\mathrm{1}} \\ $$

Commented by mr W last updated on 23/Apr/19

it is (((1/x)/(x−1)))^x =(((((x−1)/x)))/(((x/(x−1)))^(x−1) ))=(((1−(1/x)))/(((x/(x−1)))^(x−1) )) ((x/(x−1)))^(x−1) =(((x−1+1)/(x−1)))^(x−1) =(1+(1/(x−1)))^(x−1)

$${it}\:{is}\:\left(\frac{\frac{\mathrm{1}}{{x}}}{{x}−\mathrm{1}}\right)^{{x}} =\frac{\left(\frac{{x}−\mathrm{1}}{{x}}\right)}{\left(\frac{{x}}{{x}−\mathrm{1}}\right)^{{x}−\mathrm{1}} }=\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)}{\left(\frac{{x}}{{x}−\mathrm{1}}\right)^{{x}−\mathrm{1}} } \\ $$$$\left(\frac{{x}}{{x}−\mathrm{1}}\right)^{{x}−\mathrm{1}} =\left(\frac{{x}−\mathrm{1}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{{x}−\mathrm{1}} =\left(\mathrm{1}+\frac{\mathrm{1}}{{x}−\mathrm{1}}\right)^{{x}−\mathrm{1}} \\ $$

Commented by maxmathsup by imad last updated on 23/Apr/19

let A(x) =(((x−1)/x))^x ⇒A(x)=(1−(1/x))^x ⇒for x>0 ln(A(x)) =xln(1−(1/x)) but we have ln^′ (1−u) =−(1/(1−u)) =−(1+u +o(u^2 )) ⇒ ln(1−u) =−u−(u^2 /2) +o(u^3 ) ⇒ln(1−(1/x))=−(1/x) −(1/(2x^2 )) +o((1/x^3 )) (x→+∞) ⇒ xln(1−(1/x))=−1−(1/(2x)) +o((1/x^2 )) ⇒lim_(x→+∞) xln(1−(1/x))=−1 ⇒ lim_(x→+∞) ln(A(x))=−1 ⇒ lim_(x→+∞) A(x) =e^(−1) =(1/e)

$${let}\:{A}\left({x}\right)\:=\left(\frac{{x}−\mathrm{1}}{{x}}\right)^{{x}} \:\Rightarrow{A}\left({x}\right)=\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)^{{x}} \:\:\:\Rightarrow{for}\:{x}>\mathrm{0}\: \\ $$$${ln}\left({A}\left({x}\right)\right)\:={xln}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)\:\:\:{but}\:{we}\:{have}\:{ln}^{'} \left(\mathrm{1}−{u}\right)\:=−\frac{\mathrm{1}}{\mathrm{1}−{u}}\:=−\left(\mathrm{1}+{u}\:+{o}\left({u}^{\mathrm{2}} \right)\right)\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}−{u}\right)\:=−{u}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:+{o}\left({u}^{\mathrm{3}} \right)\:\Rightarrow{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)=−\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\:+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)\:\left({x}\rightarrow+\infty\right)\:\Rightarrow \\ $$$${xln}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)=−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{x}}\:+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:\Rightarrow{lim}_{{x}\rightarrow+\infty} {xln}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)=−\mathrm{1}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow+\infty} \:{ln}\left({A}\left({x}\right)\right)=−\mathrm{1}\:\Rightarrow\:{lim}_{{x}\rightarrow+\infty} \:{A}\left({x}\right)\:={e}^{−\mathrm{1}} \:=\frac{\mathrm{1}}{{e}} \\ $$$$ \\ $$

Answered by tanmay last updated on 23/Apr/19

another way t=(1/x) lim_(t→0) (1−t)^(1/t) y=lim_(t→0) (1−t)^(1/t) lny=lim_(t→0) ((ln(1−t))/(−t))×−1 lny=−1 y=e^(−1) =(1/e)

$${another}\:{way} \\ $$$${t}=\frac{\mathrm{1}}{{x}} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{{t}}} \\ $$$${y}=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{{t}}} \\ $$$${lny}=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{ln}\left(\mathrm{1}−{t}\right)}{−{t}}×−\mathrm{1} \\ $$$${lny}=−\mathrm{1} \\ $$$${y}={e}^{−\mathrm{1}} =\frac{\mathrm{1}}{{e}} \\ $$

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