Evaluate $ \lim_{x\to\infty} \frac{\ln(1+e^x)}{x} $

Question

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?

Answer 1

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 $x\rightarrow -\infty$: $$x=\ln y \Rightarrow \lim_{x\rightarrow -\infty}\frac{\ln (1+e^x)}{x} = \lim_{y\rightarrow 0^+}\frac{\ln (1+y)}{\ln y} = 0$$

Answer 2

Hint: Write $$\ln(1+e^x)=\ln\left(e^x\left(1+\frac{1}{e^x}\right)\right)=x+\ln\left(1+\frac{1}{e^x}\right)$$

Answer 3

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.

Answer 4

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

$$\ln \left(\frac{1}{e^x}+1\right)\to \ln 1=0$$