Question Number 159549 by mathlove last updated on 18/Nov/21
Answered by Ar Brandon last updated on 18/Nov/21
$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{3}^{{x}} \right)}{\mathrm{ln}\left(\mathrm{1}+\mathrm{2}^{{x}} \right)}=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{3}^{{x}} }{\mathrm{2}^{{x}} }=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\mathrm{0} \\ $$