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

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

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

$$letf\left(x\right)={(1+\frac{1}{x})}^{-\frac{1}{lnx}}\Rightarrow f\left(x\right)=e^{-\frac{1}{lnx}ln(1+\frac{1}{x})}changement\frac{1}{x}=tgive$$$$f\left(x\right)=g\left(t\right)=e^{-\frac{1}{-lnt}ln(1+t)}=e^{\frac{ln(1+t)}{lnt}}(x\to 0^{+}\Rightarrow t\to +\mathrm{\infty })\Rightarrow$$$$g\left(t\right)=e^{\frac{ln\left(t\right)+ln(1+\frac{1}{t})}{lnt}}=e^{1+\frac{ln(1+\frac{1}{t})}{lnt}}\to e(t\to +\mathrm{\infty })\Rightarrow$$$${lim}_{x\to 0^{+}}f\left(x\right)=e$$

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

$$thankyousir$$

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