A function $f$ such that $f(x)$ increases from $0$ to $1$ when $x$ increases from $0$ to infinity? I am looking for a function f(x) with a value range of [0,1].
f(x) should increase from 0 to 1 while its parameter x increases from 0 to +infinity.
f(x) increases very fast when x is small, and then very slow and eventually approach 1 when x is infinity.
Here is a figure. The green curve is what I am looking for:

Thanks.
It would be great if I can adjust the slope of the increase. Although this is not a compulsory requirement.  
 A: $f(x) = 1 - \exp(-x/\epsilon)$ for $\epsilon > 0$ small will do.
A: I think this should work well for your purposes:
$$
f(x) = \frac{x}{x + a}
$$
Where $a$ can be any number bigger than $0$.  The smaller $a$ is, the sharper the increase will be.
ADDENDUM: if you want to extend this to an odd (and continuously differentiable) function, simply take
$$
f(x) = \frac{x}{|x| + a}
$$
A: Some time ago I had an interest in such a function, but with the added requirement that it should map negative values of $x$ onto the range $(-1,0)$, asymptotically approaching $-1$
as $x$ goes to $-\infty$.
In fact, I wanted $f(-x) = -f(x)$.
I came up with something like this:
$$ f(x) = \frac{x}{\sqrt{a^2 + x^2}} .$$
This has slope $\frac{1}{|a|\,}$ at $x = 0$.
A: 
Simple version:
  $$f(x)=1-\mathrm e^{-a\sqrt{x}}\qquad (a\gt0)$$

Slightly more elaborate version:$$f(x)=1-\mathrm e^{-a\sqrt{x}-bx}\qquad (a\gt0,\ b\geqslant0)$$
Every such function fits every requisite in the question, including the infinite slope at $0$. The parameter $a$ can help to tune the increase near $0$. The parameter $b$ can help to tune the increase near $+\infty$. To get even quicker convergence to $1$ when $x\to+\infty$, one can replace $-bx$ in the exponent by $-bx^n$ for some $b\gt0$ and $n\gt1$.
Once one understands the principle, a host of other solutions springs to mind.
A: Try $f(x) = 1-e^{-x^2}$ , which will be 0 for x = 0 and approach 1 rapidly for big x.
A: Here are two simple functions with slope $k$ at $x=0$, for some $k>0$:
$$f(x) = 1-e^{-kx}$$
and
$$g(x) = \frac{2}{\pi}\arctan\left(\frac{\pi}{2}kx\right)$$
The first of these approaches $1$ more quickly than the second.
A: One I use very often is :
$$ \tanh x = \frac{\sinh x}{\cosh x} = \frac {e^x - e^{-x}} {e^x + e^{-x}} = \frac{e^{2x} - 1} {e^{2x} + 1} = \frac{1 - e^{-2x}} {1 + e^{-2x}}$$
The increase at the begining around 0 is "only" linear but can do the work. and you can choose the slope at $x \mapsto 0$

A: I'm surprised no one has mentioned erf(x) aka the "Error Function", defined roughly as the normalized area under the bell curve as a function of the upper limit of integration. For x > 1, it satisfies your requirement and also has a slope that is easily controlled (and made arbitrarily large at 0) by the width of your Gaussian.
