Find $f(x)$ and $\lim_{x\to\infty} f(x)^x$ for $f'(0)=1$, $-1
Problem. Find $f(x)$ and $\lim_{x\to\infty} f(x)^x$ for $f'(0)=1$, $-1<f(x)<1$, and
$$f(x+y)=\frac{f(x)+f(y)}{1+f(x)f(y)}
\quad\text{for all}\quad x\in\mathbb{R}.$$
Here were my findings for the following function:

*

*$f(0)=0$:

Proof: Putting $y=0$ and solving we have:
$f(0)\cdot(f(x)-1)\cdot (f(x)+1)=0$
If $f(0)\neq 0 $, then $f(x)=1$ or $f(x)=-1$ for all $x$, then $f(x)=c$, for this case so $f'(x)=0$ for any $x$, but our condition was $f'(0)=1$ hence $f(0)$ needs to be $0$.


*$f(x)$ is odd:

Putting $y=-x$ in the original equation we have: $f(0) =f(x)+f(-x)$, hence $f(x)$ is odd.
I wasn't able to move forward with the two findings. My intuition said that it is asking
$\lim_{x \to \infty} f(x)^x$ for  $-1<f(x)<1$, so its probably trying to hint me for a $1^\infty$ limit form. So maybe $f(x)$ is increasing, but I wasn't able to prove it.
Similarly to prove $f(x)$ is increasing, there was a possible need to prove $f(x)$ is differentiable at all points, and I wasn't able to this as well, I was thinking there is probably a way to get  a form of $f'(x)=g(x)$ from the original functional equation, but wasn't successful in trying to bring it to that form.
 A: $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h\\=\lim_{h\to0}\frac{\frac{f(x)+f(h)}{1+f(x)f(h)}-f(x)}h\\=\lim_{h\to0}\frac{f(x)+f(h)-f(x)-f^2(x)f(h)}{h(1+f(x)f(h))}\\=\lim_{h\to0}\frac{f(h)(1-f^2(x))}{h(1+f(x)f(h))}\\=\lim_{h\to0}\frac{f(h)-f(0)}{h}\frac{1-f^2(x)}{1+f(x)f(h)}(\text{as }f(0)=0)\\=f'(0)(1-f^2(x))$$
Thus, $f'(x)=1-f^2(x)$
Solving this differential equation, we get $$f(x)=\frac{1-e^x}{1+e^x}$$
A: Differentiating the given equation w.r.t $y$ keeping $x$ as constant and then putting $y=0$, we get $$f'(x)=1-f^2(x)$$
Solving this differential equation, we get $$f(x)=\frac{1-e^x}{1+e^x}$$
A: It is possible to get around differentiability, but it may feel a bit of cheating in the sense that the premises come out of the blue.
So notice first that for $g(x)=\frac{1-x}{1+x}$ then $g'(x)=-\frac 2{(1+x)^2}<0$
$g\searrow$ and $g(0)=1$ and $\lim\limits_{x\to+\infty}g(x)=-1$
Therefore we can set $f(x)=\dfrac{1-u(x)}{1+u(x)}$ with $u(x)>0$ and $f$ verifies $f(x)\in(-1,1)$.
Reporting in the equation $f(x+y)-\frac{f(x)+f(y)}{1+f(x)f(y)}$ gives us:
$$2\frac{u(x)u(y)-u(x+y)}{(1+\underbrace{u(x)u(y)}_{>0})(1+\underbrace{u(x+y)}_{>0})}=0$$
By positivity of $u$ the denominator is non-zero, so it should be the numerator which cancels:
$$u(x+y)=u(x)u(y)$$
And assuming only continuity of $u$ then it is an exponential, i.e. $u(x)=e^{ax}$, the differentiability comes out as a consequence.
Applying now $f'(0)=1$ gives $a=1$.
The question of the limit in infinity of $f(x)^x$ is ill-posed because $f(x)\to -1$ and the power of negative numbers is generally undefined for real exponents (at most you can give it a sense for numbers $x\in\mathbb Q_{odd}$, irreducible rationals $\frac pq$ with $q$ odd. But even though this would oscillate between $-1$ and $1$ and there is no limit).
