How to compute $\lim_{x \rightarrow \infty}\dfrac{\ln(1+e^{\alpha x})}{\ln(1+e^{\beta x})}$?

Question

I need to compute $\lim_{x \rightarrow \infty}\dfrac{\ln(1+e^{\alpha x})}{\ln(1+e^{\beta x})}$, where $α > 0$ and $β > 0$

Answer 1

Use the log functional equation.

$$\log(1+e^{\alpha x})=\log(e^{\alpha x})+\log(1+e^{-\alpha x})$$ so that

$$\lim_{x\to\infty}\dfrac{\ln(1+e^{\alpha x})}{\ln(1+e^{\beta x})}=\lim_{x\to\infty}\dfrac{\alpha x+\ln(1+e^{-\alpha x})}{\beta x+\ln(1+e^{-\beta x})}$$

The log terms now go to $0$ since $\alpha,\beta>0$, and we are left with ${\alpha\over\beta}$ as our limit.

Answer 2

Write the function as $$f(x) = \dfrac{\ln(1+e^{\alpha x})}{\ln(1+e^{\beta x})} = \dfrac{\ln(e^{\alpha x}(1+e^{-\alpha x}))}{\ln(e^{\beta x}(1+e^{-\beta x}))} = \dfrac{\alpha x + \ln(1+e^{-\alpha x})}{\beta x + \ln(1+e^{-\beta x})}$$ Since, $\alpha, \beta, x$ are all non-negative, we have $$\dfrac{\alpha x}{\beta x + \ln(2)} \leq f(x) \leq \dfrac{\alpha x + \ln(2)}{\beta x}$$ Now you should be able to compute the desired limit.