Calculate the limit of $\left(\frac{f(x)}{x}\right)^{1/x}$ Let $f(x),f:R\to R$ be a non-constant continuous function such that 
$$\left(e^x-1\right)f(2x)=\left(e^{2x}-1\right)f(x)\,.$$If $f'(0)=1$, then
$$\lim_{x\to \infty}\left(\frac{f(x)}{x}\right)^{1/x}$$
My approach: 
Well I just dont get the idea of how to start, I separated the $f(x)$ terms and tried to solve but got nowhere.
Any help will be appreciated, thanks.
 A: Rewrite the equation as
$$
f(2x) = \frac{e^{2x} - 1}{e^x - 1}f(x) = (e^x + 1)f(x)
$$
Let $L = \lim_{x \to \infty} \left(\frac{f(x)}{x}\right)^{1/x}$, supposing that this  limit exists.  Note that
$$
L = \lim_{x \to \infty} \left(\frac{f(x)}{x}\right)^{1/x} = 
\lim_{x \to \infty} \left(\frac{f(2x)}{2x}\right)^{1/(2x)} = 
\lim_{x \to \infty} \left(\frac{e^{x} + 1}{2}\frac{f(x)}{x}\right)^{1/(2x)} = \\
\sqrt{
\lim_{x \to \infty} \left(\frac{e^{x} + 1}{2}\right)^{1/x}  \lim_{x \to \infty}\left(\frac{f(x)}{x}\right)^{1/x}} = 
\sqrt{eL}
$$
Conclude that if the limit exists and is non-zero, it must be the case that $L = e$.
The fact that $f'(0) = 1$ tells us that
$$
\lim_{x \to 0} \frac{f(x) - f(0)}{x} = 1
$$
but I'm not sure how this is relevant.

Note: Let $x_0 > 0$ be such that $f(x_0) \neq 0$. 
Because $f(2x)/f(x) = e^x + 1$, we can conclude that (again, assuming the limit exists)
$$
L = \lim_{n \to \infty} \left(\frac{f(2^nx_0)}{2^nx_0}\right)^{1/(2^nx_0)} \geq\\
\lim_{n \to \infty} \left(\frac{[e^{x_0} + 1]^{2^n}f(x_0)}{2^nx_0}\right)^{1/(2^nx_0)}
$$
So it suffices to show that
$$
\lim_{m \to \infty} \left(\frac{[e^{x_0} + 1]^{m}}{m}\right)^{1/m} = e^{x_0} + 1 > 0
$$
A: Suppose that
$$
L=\lim_{x\to\infty}\left(\frac{f(x)}x\right)^{1/x}
$$
exists. Then
$$
\begin{align}
L&=
\lim_{x\to\infty}\left[\left(\frac{f(2x)}{2x}\right)^{1/(2x)}\frac{\left(\frac{f(2x)}{2x}\right)^{1/(2x)}}{\left(\frac{f(x)}x\right)^{1/x}}\right]\\
&=\lim_{x\to\infty}\frac{\left(\frac{f(2x)}{2x}\right)^{1/x}}{\left(\frac{f(x)}x\right)^{1/x}}\\
&=\lim_{x\to\infty}\left(\frac12\frac{e^{2x}-1}{e^x-1}\right)^{1/x}\\[9pt]
&=\lim_{x\to\infty}\left(\frac{e^x+1}2\right)^{1/x}\\[12pt]
&=e
\end{align}
$$

Note that
$$
\begin{align}
\frac{\frac{f\left(2^{n+1}\right)}{2^{n+1}}}{\frac{f\left(2^n\right)}{2^n}}
&=\frac12\frac{e^{2^{n+1}}-1}{e^{2^n}-1}\\
&=\frac{e^{2^n}+1}2
\end{align}
$$
This means that $\frac{f\left(2^n\right)}{2^n}$ is increasing for all $n\in\mathbb{Z}$. Since $f'(0)=1$, we get
$$
\begin{align}
\lim_{n\to-\infty}\frac{f\!\left(2^n\right)}{2^n}
&=\lim_{x\to0}\frac{f(x)}x\\[6pt]
&=1
\end{align}
$$
Therefore, for all $n\in\mathbb{Z}$,
$$
\left(\frac{f(2^n)}{2^n}\right)^{1/2^n}\ge1
$$
This means that, should it exist,
$$
\lim_{x\to\infty}\left(\frac{f(x)}{x}\right)^{1/x}\ge1
$$
In particular, the limit is not $0$.
A: From the equation $$f(x)=\frac{e^x-1}{e^{\frac{x}{2}}-1}f(\frac{x}{2})=\frac{e^x-1}{e^{\frac{x}{2}}-1}\frac{e^{\frac{x}{2}}-1}{e^{\frac{x}{2^2}}-1}f(\frac{x}{2^2})$$
By induction $$f(x)=\frac{e^x-1}{e^{\frac{x}{2^n}}-1}f(\frac{x}{2^n})$$
From $f(2x)=(e^x+1)f(x)$, we can see that $f(0)=0$. Thus letting $n \to \infty$ on both sides of the above equation we have $$f(x)=(e^x-1)\lim_{n \to \infty}\frac{f(\frac{x}{2^n})}{e^{\frac{x}{2^n}}-1}=e^x-1$$
Then you see that the limit is $e$. 
