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: How about using $\displaystyle e^x < 1+e^x < 2e^x$, for $x > 0$, taking $\log$ and apply the Squeeze Theorem to conclude the required limit is $1$. (which is essentially same as the other hints).
For $x\to\infty$
$$\log(1+e^x)\sim\log(e^x)=x$$
Therefore, the limit can be easily simplified
$$\lim_{x\to\infty}\frac{\log(1+e^x)}{x}=\lim_{x\to\infty}\frac{x}{x}=1$$