# Finding $f(x)$ from functional equation

Let $f: \mathbb R \to \mathbb R$ be a non constant continuous function such that $(e^x-1)f(2x)= (e^{2x}-1)f(x)$. 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?

$e^{\frac12}$

• A quick look tells me that $f(x)=e^x-1$. I don't know if that is the only possibility for $f$ – randomgirl Jun 9 '18 at 16:28
• Pretending that is the only $f$.... you should get your spoiler you posted. – randomgirl Jun 9 '18 at 16:42
• @randomgirl how? – Archer Jun 9 '18 at 16:49
• m.wolframalpha.com/input/… – randomgirl Jun 9 '18 at 16:53
• @randomgirl I wanted to know the method instead – Archer Jun 9 '18 at 16:54

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: $$\lim_{x\to 0} \left(\frac{f(x)}{x}\right)^{1/x} = \lim_{x\to 0} \exp\left[\frac{1}{x}\log\left(\frac{e^x-1}{x}\right)\right] = \lim_{x\to 0} \exp\left[\frac{1}{x}\log\left(1+\frac{x}{2}\right)\right] = e^{1/2}.$$

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 $$\lim_{x\to 0}\left(\frac{e^x-1}x\right)^{1/x} = e^{\lim_{x\to 0}\frac{\ln (e^x-1) - \ln x}{x}} = e^{1/2}$$

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.

• I think you are assuming that the limit exists. – Rigel Jun 9 '18 at 17:15
• Yeah! That is a loophole in this proof. – Lord KK Jun 9 '18 at 17:26
• I believe it should be $$\lim_{x \to 0}\left (\frac {f(x)}x \right)^{1/x} =\lim_{x \to 0}\left (\frac {f(2x)}{2x} \right)^{\frac1{2x}} =\lim_{x \to 0}\left (\frac {(e^{2x}-1)f(x)}{(e^{x}-1)2x} \right)^{\frac1{2x}}$$ and the limit $$\lim_{x\to0} \left(2\cdot \frac{e^{2x}-1}{e^x-1}\right)^{\frac1{2x}}$$ does not exist. – mechanodroid Jun 9 '18 at 17:44
• @mechanodroid Thanks for the correction. Edited. – Lord KK Jun 9 '18 at 17:54