Evaluate $ \lim_{x\to\infty} \frac{\ln(1+e^x)}{x} $ I have intuitively guessed that the answer should be 1 as $x\to +\infty$. How do I formally prove this? Also how do I evaluate the limit as $x\to -\infty $.
UPDATE: Based on the answers below, I have managed to do the following.
$$
\lim_{x\to\infty} \frac{\ln (1+e^x)}{x} \\
= \lim_{x\to\infty} \frac{\ln e^x(\frac{1}{e^x}+1)}{x} \\
= \lim_{x\to\infty} \frac{x +\ln (\frac{1}{e^x}+1)}{x} \\
= \lim_{x\to\infty} 1+ \ln (1+\frac{1}{e^x})^\frac{1}{x} \\
$$
How do I proceed proceed from here?
 A: Hint: Write $$\ln(1+e^x)=\ln\left(e^x\left(1+\frac{1}{e^x}\right)\right)=x+\ln\left(1+\frac{1}{e^x}\right)$$
A: Note that
$$ \frac{\ln(1+e^x)}{x}=\frac{\ln e^x+\ln(1+1/e^x)}{x}=\frac{x+\ln(1+1/e^x)}{x}$$
Based on your work, from here
$$...= \lim_{x\to\infty} \frac{x +\ln (\frac{1}{e^x}+1)}{x}=\lim_{x\to\infty} 1+ \frac{ \ln (\frac{1}{e^x}+1)}{x}=1$$
indeed
$$\ln \left(\frac{1}{e^x}+1\right)\to \ln 1=0$$
A: For $x\rightarrow +\infty$:
$$1 = \frac{\ln e^x}{x} \leq \frac{\ln (1+e^x)}{x} \leq \frac{\ln (2e^x)}{x}=1+\frac{\ln 2}{x} \stackrel{x\rightarrow +\infty}{\longrightarrow}1$$
For $x\rightarrow -\infty$:
$$x=\ln y \Rightarrow \lim_{x\rightarrow -\infty}\frac{\ln (1+e^x)}{x} = \lim_{y\rightarrow 0^+}\frac{\ln (1+y)}{\ln y} = 0$$
A: Since you would get $\frac{\infty}{\infty}$
You can use l'hopital's rule:
$ \lim_{x\to\infty} \frac{\ln(1+e^x)}{x}=\frac{e^x}{1+e^x}=\frac{1}{e^{-x}+1}=1$
Formal proof of this relies on the proof of this case for l'hopital's rule.
Add on: for $-\infty$, solution should be trivial, treat $-\infty$ as an extended real number and substitute.
