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Question Number 174331 by cortano1 last updated on 30/Jul/22

Commented by kaivan.ahmadi last updated on 30/Jul/22

$$\stackrel{hop}{\sim }li\underset{x\to \mathrm{\infty }}{m}\frac{\frac{3e^{3x}}{2+e^{3x}}}{\frac{2e^{2x}}{3+e^{2x}}}=li\underset{x\to \mathrm{\infty }}{m}\frac{3e^{3x}(3+e^{2x})}{2e^{2x}(2+e^{3x})}$$$$$$$$=li\underset{x\to \mathrm{\infty }}{m}\frac{3e^x(3+e^{2x})}{2(2+e^{3x})}\stackrel{hop}{\sim }$$$$li\underset{x\to \mathrm{\infty }}{m}\frac{9e^x+9e^{3x}}{6e^{3x}}=\frac{3}{2}$$$$$$

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