Which function satisfies these constraints? 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}
 A: Differentiating $f(-x) = f(x) \exp(-x)$, we have $-f'(-x) = (f'(x) - f(x)) \exp(-x)$.  If $f$ is to be monotone increasing we'll need this to be nonpositive, so $f'(x) - f(x) \le 0$.  You could take
$$ f(x) = \cases{1 + x & for $x \ge 0$\cr
                 (1-x) \exp(x) & for $x < 0$\cr}$$
EDIT:  But there is an analytic solution, namely 
$$ f(x) = \frac{x e^x}{e^x-1} $$
A: Lots of functions work, e.g.
$$f(x) = \left\{\begin{array}{cc} x + 1 & x \ge 0 \\ e^x ({1 - x}) & x < 0\end{array}\right.$$

I came up with this in two steps:


*

*Hoped that $f(x) = x$ for positive $x$ works. It doesn't, because it's zero at zero.

*Fix this by adding $1$, or any positive number to it. 
A: 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
$$
\begin{align}
f(x)
&=\frac{x/2\,e^{x/2}}{\sinh(x/2)}\\[6pt]
&=\frac x2\left(1+\coth\left(\frac x2\right)\right)
\end{align}
$$
A: $f(x)=e^{x}(-x+1)$ when $x<0$ and $f(x)=x+1$ when $x\geq 0$.
