TITLE: Which function satisfies these constraints?
QUESTION [6 upvotes]: What is the simplest monotone increasing, continuous function that satisfies these constraints?
\begin{align}
\frac{f(x)}{f(-x)} &= e^x \\
\lim_{x\to\infty} \frac{f(x)}{x} &= 1
\end{align}
I guess this implies that
\begin{align}
\lim_{x\to-\infty} \frac{f(x)}{-xe^{x}} &= 1?
\end{align}

REPLY [4 votes]: The odd part of $\log(f(x))$ is
$$
\begin{align}
\frac12\left(\vphantom{\frac12}\log(f(x))-\log(f(-x))\right)
&=\frac12\log\left(\frac{f(x)}{f(-x)}\right)\\
&=\frac x2
\end{align}
$$
Therefore, $f(x)=e^{\frac x2+g(x)}$ where $g(x)$ is an even function.
We also want $\lim\limits_{x\to\infty}\left(\frac x2+g(x)-\log(x)\right)=0$. Thus, let $g(x)=\log\left(\frac{x/2}{\sinh(x/2)}\right)$ and get
$$
\begin{align}
f(x)
&=\frac{x/2\,e^{x/2}}{\sinh(x/2)}\\[6pt]
&=\frac x2\left(1+\coth\left(\frac x2\right)\right)
\end{align}
$$