Question Number 196337 by SaRahAli last updated on 22/Aug/23
Answered by MM42 last updated on 22/Aug/23
$${lim}_{{x}\rightarrow\infty} \:\frac{{ln}\left(\mathrm{1}+\frac{{a}}{{x}}\right)}{\frac{\mathrm{1}}{{x}}}\:\overset{\frac{\mathrm{1}}{{x}}=\:{u}} {=}{lim}_{{u}\rightarrow\mathrm{0}} \:\frac{{ln}\left(\mathrm{1}+{au}\right)}{{u}} \\ $$$${hop}={lim}_{{u}\rightarrow\mathrm{0}} \:\frac{\frac{{a}}{\mathrm{1}+{au}}}{\mathrm{1}}\:={a}\:\checkmark \\ $$