Question Number 202466 by Rydel last updated on 27/Dec/23
$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{x}\sqrt{\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)}}{\mathrm{1}+{e}^{{x}−\mathrm{3}} } \\ $$
Answered by Mathspace last updated on 27/Dec/23
$$={lim}_{{x}\rightarrow+\infty} {xe}^{−{x}+\mathrm{3}} \sqrt{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}/\mathrm{1}+{e}^{−{x}+\mathrm{3}} \\ $$$$={lim}_{{x}\rightarrow+\infty} {xe}^{−{x}+\mathrm{3}} \sqrt{\mathrm{2}{lnx}+{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)} \\ $$$$={lim}_{{x}\rightarrow+\infty} {xe}^{−{x}+\mathrm{3}} \sqrt{\mathrm{2}{lnx}} \\ $$$$=\mathrm{0}\:\:{the}\:{boss}\:{here}\:{is}\:{e}^{−{x}} \\ $$