Question Number 79679 by ~blr237~ last updated on 27/Jan/20
$${Find}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} {ln}\left(\mathrm{1}+{x}\right)−{x} \\ $$
Commented by mathmax by abdo last updated on 27/Jan/20
$${let}\:{f}\left({x}\right)={x}^{\mathrm{2}} {ln}\left(\mathrm{1}+{x}\right)−{x}\:\Rightarrow{f}\left({x}\right)={x}^{\mathrm{2}} {ln}\left({x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\right)−{x} \\ $$$$={x}^{\mathrm{2}} {lnx}\:+{x}^{\mathrm{2}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)−{x}\sim{x}^{\mathrm{2}} {lnx}+{x}^{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}}\right)−{x}\:\Rightarrow \\ $$$${f}\left({x}\right)\sim{x}^{\mathrm{2}} {lnx}\:\Rightarrow{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=+\infty \\ $$$$ \\ $$
Commented by ~blr237~ last updated on 27/Jan/20
$${thanks}\:{sir} \\ $$