\[\ln(x+1) = \ln \Big(x \times (1+\frac1x)\Big) = \ln x+\ln\big(1+{\frac1x}\big) = \ln x + \frac 1x + o(\frac1x)\] Donc \[\frac{\ln(x+1)}{\ln x} = 1 + \frac1{x\ln x} + o(\frac1{x\ln x}).\] Ainsi \[\begin{align*} \left(\frac{\ln(x+1)}{\ln x}\right)^x & = \exp\left( x \ln \left( \frac{\ln(x+1)}{\ln x}\right) \right) \\ & = \exp\left( x \ln \left(1 + \frac1{x\ln x} + o(\frac1{x\ln x})\right) \right) \\ & = \exp\left( x \left(\frac1{x\ln x} + o(\frac1{x\ln x})\right) \right) \\ & = \exp\left( \frac1{\ln x} + o(\frac1{\ln x})\right) \\ & = 1 + \frac1{\ln x} + o(\frac1{\ln x}) \\ \end{align*}\] On en déduit immédiatement que \[\lim_{x\to+\infty}\left(\frac{\ln(x+1)}{\ln x}\right)^x = 1\] et que lorsque \(x\to + \infty\) \[\left(\frac{\ln(x+1)}{\ln x}\right)^x - 1 \sim \frac1{\ln x}.\]

exo7 2657