Find functions $f:\mathbb{R}\to[0,1]$ such that: $\lim_{x \to +\infty} \frac{f(x)^2}{f(2 x)}=1$ Find functions $f:\mathbb{R}\to[0,1]$ that satisfy: $$\lim_{x \to +\infty} \frac{f(x)^2}{f(2 x)}=1,$$
$$f'(x)\leq 0 \, \forall x,$$
and
$$\lim_{x \to +\infty} f(x)=0.$$
$$\lim_{x \to -\infty} f(x)=1.$$
I tried to find one without success.

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: $$f(x) = \frac{1}{e^x+1}$$
