How to evaluate $\lim\limits_{x \to \infty}\dfrac{\ln{\left(1+e^x\right)}}{x}$ without using L'Hôpital's rule?

Question

The given limit is from an exercise from the book Differential and Integral Calculus by Piskunov:

$$\lim\limits_{x \to \infty}\dfrac{\ln{\left(1+e^x\right)}}{x}$$

The answer in the book is $1$ as $x \to +\infty$ but $0$ as $x \to -\infty$.

Answer 1

The case $x\rightarrow-\infty$ is easy: $e^x\rightarrow0$, so $\ln(1+e^x)\rightarrow\ln(1+0)=0$ and $\frac{1}{x}\rightarrow0$

For $x\rightarrow+\infty$, notice that $$\ln (1+e^x)=\ln (1+e^x)+x-x=\ln (1+e^x)-\ln e^x+x=\ln\left(\frac{1+e^x}{e^x}\right)+x =\ln (e^{-x}+1)+x$$ Notice that $\ln (e^{-x}+1)\rightarrow0$, so $\cdots$, can you continue from here?

Answer 2

Well $x=\ln e^x$ so \begin{align} \lim\limits_{x \to \infty}\dfrac{\ln{\left(1+e^x\right)}}{x} &= \lim\limits_{x \to \infty}\dfrac{\ln{\left(1+e^x\right)}}{\ln e^{x}}\\ &= \lim\limits_{x \to \infty}\dfrac{\ln e^x{\left(1+e^{-x}\right)}}{\ln e^x}\\ & =\lim\limits_{x \to \infty}\left(\dfrac{\ln {\left(1+e^{-x}\right)}}{\ln e^x}+1\right)\\ &\ldots \end{align}

Answer 3

$z:=e^x$, and cosider $\lim z \rightarrow \infty.$

$\dfrac{\log (z(1+1/z))} {\log z} =$

$1+\dfrac{\log (1+1/z)}{\log z};$

Take the limit.

Used: $\log (ab) =\log a +\log b$.

Answer 4

$$ \begin{align*} \lim_{x\rightarrow +\infty} \frac{\ln \left( 1+e^x \right)}{x}&=\lim_{x\rightarrow +\infty} \frac{\ln \left[ e^x\left( 1+e^{-x} \right) \right]}{x} \\ &=1+\lim_{x\rightarrow +\infty} \frac{\ln \left( 1+e^{-x} \right)}{x} \\ &=1+\lim_{x\rightarrow +\infty} \frac{e^{-x}}{x} \\ &=1+\lim_{x\rightarrow +\infty} \frac{1}{xe^x} \\ &=1 \end{align*} $$

$$ \begin{align*} \lim_{x\rightarrow -\infty} \frac{\ln \left( 1+e^x \right)}{x}=\lim_{x\rightarrow -\infty} \frac{e^x}{x}=0 \end{align*} $$