Question Number 65464 by aliesam last updated on 30/Jul/19
Commented by mathmax by abdo last updated on 30/Jul/19
$${let}\:\:{A}\left({x}\right)\:={x}\left({x}+\mathrm{1}\right){ln}\left(\frac{{x}+\mathrm{1}}{{x}}\right)−{x}\:\:\:\:{we}\:{have}\:{for}\:{x}\in{V}\left(+\infty\right) \\ $$$${ln}\left(\frac{{x}+\mathrm{1}}{{x}}\right)\:={ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\:=\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\:+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\: \\ $$$${A}\left({x}\right)\:=\left({x}^{\mathrm{2}} +{x}\right)\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\:+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\right)−{x} \\ $$$$={x}−\frac{\mathrm{1}}{\mathrm{2}}\:+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{x}}\:−{x}\:+{o}\left(\frac{\mathrm{1}}{{x}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}{x}}\:+{o}\left(\frac{\mathrm{1}}{{x}}\right)\:\Rightarrow{lim}_{{x}\rightarrow+\infty} {A}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$