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Question Number 82815 by M±th+et£s last updated on 24/Feb/20

lim_(x→0^+ ) (1/((1+(1/x))^(1/(ln(x))) ))=?

$$\underset{{x}\rightarrow\mathrm{0}^{+} } {{lim}}\:\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{\frac{\mathrm{1}}{{ln}\left({x}\right)}} }=? \\ $$

Commented by mathmax by abdo last updated on 24/Feb/20

let f(x)=(1+(1/x))^(−(1/(lnx))) ⇒f(x)=e^(−(1/(lnx))ln(1+(1/x))) changement (1/x)=t give f(x)=g(t) =e^(−(1/(−lnt))ln(1+t)) =e^((ln(1+t))/(lnt)) (x→0^+ ⇒t→+∞) ⇒ g(t)=e^((ln(t)+ln(1+(1/t)))/(lnt)) =e^(1+((ln(1+(1/t)))/(lnt))) →e (t→+∞) ⇒ lim_(x→0^+ ) f(x)=e

$${let}\:{f}\left({x}\right)=\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{−\frac{\mathrm{1}}{{lnx}}} \:\Rightarrow{f}\left({x}\right)={e}^{−\frac{\mathrm{1}}{{lnx}}{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)} \:\:{changement}\:\frac{\mathrm{1}}{{x}}={t}\:{give} \\ $$$${f}\left({x}\right)={g}\left({t}\right)\:={e}^{−\frac{\mathrm{1}}{−{lnt}}{ln}\left(\mathrm{1}+{t}\right)} \:={e}^{\frac{{ln}\left(\mathrm{1}+{t}\right)}{{lnt}}} \:\:\left({x}\rightarrow\mathrm{0}^{+} \:\Rightarrow{t}\rightarrow+\infty\right)\:\Rightarrow \\ $$$${g}\left({t}\right)={e}^{\frac{{ln}\left({t}\right)+{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right)}{{lnt}}} ={e}^{\mathrm{1}+\frac{{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right)}{{lnt}}} \:\rightarrow{e}\:\left({t}\rightarrow+\infty\right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{f}\left({x}\right)={e} \\ $$

Commented by M±th+et£s last updated on 24/Feb/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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