Question Number 73552 by Rio Michael last updated on 13/Nov/19
$${evaluate}\: \\ $$$$\underset{{x}\rightarrow+\infty} {\:{Lim}}\:{xln}\:\left(\frac{{x}+\mathrm{1}}{{x}}\right) \\ $$
Commented by mathmax by abdo last updated on 13/Nov/19
$${we}\:{have}\:{xln}\left(\frac{{x}+\mathrm{1}}{{x}}\right)={xln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\:={x}\left(\frac{\mathrm{1}}{{x}}\:+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\right)\:=\mathrm{1}+{o}\left(\frac{\mathrm{1}}{{x}}\right)\left({x}\rightarrow+\infty\right) \\ $$$$\Rightarrow{lim}_{{x}\rightarrow+\infty} \:\:{xln}\left(\frac{{x}+\mathrm{1}}{{x}}\right)\:=\mathrm{1} \\ $$
Answered by ajfour last updated on 13/Nov/19
$${L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)}{\left(\frac{\mathrm{1}}{{x}}\right)}=\:\mathrm{1} \\ $$
Commented by Rio Michael last updated on 13/Nov/19
$${thanks} \\ $$