Prove that $\lim_{x \to \infty} \frac{\log(1+e^x)}{x} = 1$ Show that
$$\lim_{x \to \infty} \frac{\log(1 + e^x)}{x} = 1$$
How do I prove this? Or how do we get this result? Here $\log$ is the natural logarithm.

Hint: Use that
$$\log(1 + e^x) = \log[e^x (1 + e^{-x})]$$

$$\cdots= \log e^x + \log(1 + e^{-x}) = x + \log(1 + e^{-x}).$$

