Finding $f(x)$ from the functional equation $(e^x-1)f(2x)= \left(e^{2x}-1\right)f(x)$ and $f'(0) =1 $ Let $f: \mathbb R \to \mathbb R$ be a non constant continuous function such that $(e^x-1)f(2x)= \left(e^{2x}-1\right)f(x)$. If $f'(0) =1 $, then $\lim_{x\to 0}\left(\dfrac {f(x)}{x}\right)^{\frac 1x}= ? $
I am trying hard to find $f(x)$ but unable to.
Attempt:
Differentiating both sides of the equation and putting $x=0$, we obtain $f(0)= 0$.
After that can we directly substitute $f(x)=$ exact $0$ and say that limit doesn't exist (which is not the answer)?
How do I go about solving it?
The answer is:

 $e^{\frac12}$

 A: Let us prove that $f(x) = e^x-1$ is the unique function satisfying the given conditions.
Namely, as already observed, differentiating the functional equation and evaluating at $x=0$ gives $f(0) = 0$.
Let us consider the function $g(x) := f(x) / (e^x - 1)$, $x\neq 0$.
Since $f(0) = 0$ and $f'(0) = 1$, there exists the limit
$$
\lim_{x\to 0} \frac{f(x)}{e^x -1} = \lim_{x\to 0} \frac{f(x)}{x}\cdot\frac{x}{e^x-1} = f'(0) = 1,
$$ 
so that $g$ can be continuously extended to all $\mathbb{R}$ by setting $g(0) = 1$.
Hence, $g$ is a continuous function satisfying
$$
g(0) = 1, \qquad g(2x) = g(x), \quad \forall x\in\mathbb{R}.
$$ 
But the only continuous function satisfying these conditions is the constant function $g \equiv 1$, i.e., $f(x) = e^x - 1$.
Namely, given $x\neq 0$ it holds
$$
g(x) = g\left( \frac{x}{2^n}\right), \qquad \forall n\in\mathbb{N},
$$
hence
$$
g(x) = \lim_{n\to +\infty} g\left( \frac{x}{2^n}\right) = g(0) = 1.
$$
Now the computation of the required limit is easy:
$$
\lim_{x\to 0} \left(\frac{f(x)}{x}\right)^{1/x}
=
\lim_{x\to 0} \exp\left[\frac{1}{x}\log\left(\frac{e^x-1}{x}\right)\right]
=
\lim_{x\to 0} \exp\left[\frac{1}{x}\log\left(1+\frac{x}{2}\right)\right]
= e^{1/2}.
$$
A: This solution only assumes $f$ is differentiable at $0$. 
Fix $x\ne 0$ and notice that
$$f(x)\left(e^{\frac{x}2} - 1\right) = f\left(2\cdot \frac{x}2\right)\left(e^{\frac{x}2} - 1\right) = f\left(\frac{x}2\right)(e^x-1)$$
so inductively we get
$$f\left(\frac{x}{2^n}\right)(e^x-1) = f(x)\left(e^{\frac{x}{2^n}} - 1\right)$$
Letting $n\to\infty$ and using continuity of $f$ at $0$, we get
$$f(0)(e^x-1) = 0 \implies f(0) = 0$$
Now for $x \ne 0$ we have
$$\frac{f(x)}{e^x-1} = \frac{f\left(\frac{x}{2^n}\right)}{e^{\frac{x}{2^n}}-1} = \frac{f\left(\frac{x}{2^n}\right)}{\frac{x}{2^n}}\frac{\frac{x}{2^n}}{e^{\frac{x}{2^n}}-1} = \frac{f\left(\frac{x}{2^n}\right) - f(0)}{\frac{x}{2^n}}\frac{\frac{x}{2^n}}{e^{\frac{x}{2^n}}-1} \xrightarrow{n\to\infty} f'(0)\cdot1 = 1$$
Therefore $f(x) = e^x-1$ for all $x \ne 0$, and by inspection we also have $f(0) = 1 = e^0-1$.
Hence $f(x) = e^x-1, \forall x \in \mathbb{R}$.
Now the desired limit is
$$\lim_{x\to 0}\left(\frac{f(x)}x\right)^{1/x} = \lim_{x\to 0}\left(\frac{e^x-1}x\right)^{1/x}$$ 
which can be computed by taking the logarithm and applying l'Hopital thrice:
$$\lim_{x\to 0}\frac{\ln (e^x-1) - \ln x}{x} \stackrel{0/0}= \lim_{x\to 0}\frac{\frac{e^x}{e^x-1} - \frac1x}{1} = \lim_{x\to 0} \frac{xe^x-e^x+1}{x(e^x-1)} \stackrel{0/0}= \lim_{x\to 0}\frac{xe^x+e^x-e^x}{xe^x+e^x-1} = \lim_{x\to 0}\frac{xe^x}{e^x(x+1)} \stackrel{0/0}= \lim_{x\to 0}\frac{xe^x+e^x}{e^x(x+1) + e^x} = \frac{0+1}{1\cdot1 + 1} = \frac12$$
Hence 
$$\lim_{x\to 0}\left(\frac{e^x-1}x\right)^{1/x} = e^{\lim_{x\to 0}\frac{\ln (e^x-1) - \ln x}{x}} = e^{1/2}$$
A: Write  $\lim_{x \to 0}\left (\frac {f(x)}x  \right)^{1/x} =\lim_{x \to 0}\left (\frac {f(2x)}{2x}  \right)^{\frac1{2x}} =\lim_{x \to 0}\left (\frac {(e^{2x}-1)f(x)}{(e^{x}-1)2x}  \right)^{\frac1{2x}} $
Thus,  if $\lim_{x \to 0}\left (\frac {f(x)}x  \right)^{1/x}=L$ 
$ \implies \lim_{x \to 0} \left ( \frac {e^{2x}-1}{2(e^{x}-1)}\right)^{\frac1{2x}}$ $= \sqrt L $
From here it should be easy.
