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:
$$\log(1+e^x) = \log(e^x) + (\log(1+e^x) - \log(e^x))=x + \log\left(\frac{1+e^x}{e^x}\right).$$
