Question Number 174331 by cortano1 last updated on 30/Jul/22
Commented by kaivan.ahmadi last updated on 30/Jul/22
$$\overset{{hop}} {\sim}\:{li}\underset{{x}\rightarrow\infty} {{m}}\:\frac{\frac{\mathrm{3}{e}^{\mathrm{3}{x}} }{\mathrm{2}+{e}^{\mathrm{3}{x}} }}{\frac{\mathrm{2}{e}^{\mathrm{2}{x}} }{\mathrm{3}+{e}^{\mathrm{2}{x}} }}={li}\underset{{x}\rightarrow\infty} {{m}}\frac{\mathrm{3}{e}^{\mathrm{3}{x}} \left(\mathrm{3}+{e}^{\mathrm{2}{x}} \right)}{\mathrm{2}{e}^{\mathrm{2}{x}} \left(\mathrm{2}+{e}^{\mathrm{3}{x}} \right)} \\ $$$$ \\ $$$$={li}\underset{{x}\rightarrow\infty} {{m}}\:\frac{\mathrm{3}{e}^{{x}} \left(\mathrm{3}+{e}^{\mathrm{2}{x}} \right)}{\mathrm{2}\left(\mathrm{2}+{e}^{\mathrm{3}{x}} \right)}\:\overset{{hop}} {\sim} \\ $$$${li}\underset{{x}\rightarrow\infty} {{m}}\frac{\mathrm{9}{e}^{{x}} +\mathrm{9}{e}^{\mathrm{3}{x}} }{\mathrm{6}{e}^{\mathrm{3}{x}} }=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$