Find functions $f:\mathbb{R}\to[0,1]$ such that: $\lim_{x \to +\infty} \frac{f(x)^2}{f(2 x)}=1$
An example is $$f(x)=e^{-x-\sqrt{x^2+1}}$$
HOW DID I FIND THIS?
The first requirement is satisfied by functions like $e^{\alpha x}$, so I thought that looking at a possible logarithm of $f$ would have been a good idea. So we need that $\log f$ has a negative asymptote at $+ \infty$
Fourth conditions say that $\log f$ must have an horizontal asymptote at $- \infty$, tending to $0$
So, I thought that an hyperbola would work. Such an hyperbola is for example $$y(y+2x)-1=0$$, which gives you $$y=-x-\sqrt{x^2+1}$$ for $y > 0$.
EDIT: I just found another simpler one: $$f(x) = \frac{1}{e^x+1}$$
Based on the constructions above we can see that any function of the form $$\frac{1}{a^{cx}+1}$$ for $a>0$ and $c \in \mathbb{R}$ will work. It would appear that you want some type of generalization of a sigmoid function (just a guess). Perhaps Gompertz functions might be a good thing to look at.
All functions $$f_a(x) = \frac{1}{2}\left(1-\tanh(ax)\right), a> 0$$ solve your requirements. Since $a$ is just a rescaling, we need only show this for $a=1$. Let $f(x) = f_a(x) = f_1(x)$. The derivative is $f'(x) = \frac{1}{2}\left(\tanh(x)^2 - 1\right) \le 0$ and $$\frac{f(x)^2}{f(2x)}=\frac{1}{2}\frac{2\cosh(x)^2-1}{\cosh(x)^2} = 1- \frac{1}{2\cosh(x)^2} < 1 $$ with a limit $1$ for $x\rightarrow \infty$. The limits of $f(x)$ for $x\rightarrow \pm \infty$ are trivial.