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Question Number 53647 by Tawa1 last updated on 24/Jan/19

lim_(π›—β†’βˆž)   (1 + (1/Ο†))^Ο†   =  e

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

Commented by maxmathsup by imad last updated on 24/Jan/19

we have (1+(1/x))^x  =e^(xln(1+(1/x)))   but  ln(1+(1/x))∼(1/x)  (xβ†’+∞) β‡’  xln(1+(1/x))∼ 1 β‡’lim_(xβ†’+∞) (1+(1/x))^x =e .

$${we}\:{have}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} \:={e}^{{xln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)} \:\:{but}\:\:{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\sim\frac{\mathrm{1}}{{x}}\:\:\left({x}\rightarrow+\infty\right)\:\Rightarrow \\ $$$${xln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\sim\:\mathrm{1}\:\Rightarrow{lim}_{{x}\rightarrow+\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} ={e}\:. \\ $$

Commented by Tawa1 last updated on 24/Jan/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Jan/19

t=(1/φ)  lim_(t→0) (1+t)^(1/t) =y(say)  lim_(t→0) ((ln(1+t))/t)=1=lny  lny=1 so y=e^1 =e

$${t}=\frac{\mathrm{1}}{\phi} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{t}\right)^{\frac{\mathrm{1}}{{t}}} ={y}\left({say}\right) \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{ln}\left(\mathrm{1}+{t}\right)}{{t}}=\mathrm{1}={lny} \\ $$$${lny}=\mathrm{1}\:{so}\:{y}={e}^{\mathrm{1}} ={e} \\ $$

Commented by Tawa1 last updated on 24/Jan/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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