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Question Number 138473 by henderson last updated on 14/Apr/21

hi ! calculate : lim_(x→∞) [((ln(x+1))/(ln (x)))]^(x ln (x)) .

$$\boldsymbol{\mathrm{hi}}\:! \\ $$$$\boldsymbol{\mathrm{calculate}}\::\:\underset{{x}\rightarrow\infty} {{lim}}\:\left[\frac{{ln}\left({x}+\mathrm{1}\right)}{{ln}\:\left({x}\right)}\right]^{{x}\:{ln}\:\left({x}\right)} . \\ $$

Answered by mathmax by abdo last updated on 14/Apr/21

let f(x)=(((ln(x+1))/(lnx)))^(xln(x)) ⇒f(x)=e^(xln(x)ln(((ln(x+1))/(lnx)))) and ln(((ln(x+1))/(lnx)))=ln(((lnx+ln(1+(1/x)))/(lnx)))=ln(1+((ln(1+(1/x)))/(lnx))) ∼ln(1+(1/(xlnx)))∼(1/(xlnx)) ⇒xlnxln(((ln(x+1))/(ln)))∼ 1 ⇒lim_(x→+∞) f(x)=e

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\left(\frac{\mathrm{ln}\left(\mathrm{x}+\mathrm{1}\right)}{\mathrm{lnx}}\right)^{\mathrm{xln}\left(\mathrm{x}\right)} \:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{e}^{\mathrm{xln}\left(\mathrm{x}\right)\mathrm{ln}\left(\frac{\mathrm{ln}\left(\mathrm{x}+\mathrm{1}\right)}{\mathrm{lnx}}\right)} \\ $$$$\mathrm{and}\:\mathrm{ln}\left(\frac{\mathrm{ln}\left(\mathrm{x}+\mathrm{1}\right)}{\mathrm{lnx}}\right)=\mathrm{ln}\left(\frac{\mathrm{lnx}+\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)}{\mathrm{lnx}}\right)=\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)}{\mathrm{lnx}}\right) \\ $$$$\sim\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{xlnx}}\right)\sim\frac{\mathrm{1}}{\mathrm{xlnx}}\:\Rightarrow\mathrm{xlnxln}\left(\frac{\mathrm{ln}\left(\mathrm{x}+\mathrm{1}\right)}{\mathrm{ln}}\right)\sim\:\mathrm{1}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{e} \\ $$

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