Looking for a function... I'm looking for a monotonic increasing function $f(x)$ defined on $[0,\infty)$ so that $f(0)=0$ and with $f'(x)$ approaching $0$ as $x$ approaches $0$, and which approaches 1 as $x \rightarrow \infty$.  
The function might look roughly like the figure below, but that particular function is just a shifted and scaled $s$-function.  In particular, $f(0)\neq 0$, nor does $f'(x)$ approach $0$ at the origin, which are two properties I'd like to have.
Any ideas? 

 A: You could try$$f(x)=\frac{x^2}{x^2+1}.$$
A: $f(x) = 1 - e^{-a x^2}$ with $a \gt 0$ satisfies the conditions $f(0)=f'(0)=0$, $\lim_{x \to \infty} f(x) = 1$.
A: Let $g:[0,\infty)\rightarrow[0,\infty)$ be a continuous function with $g(0)=0$ and $\int_0^\infty g(s)ds = M < \infty$. Then, simply define $f(x) := \frac{1}{M}\int_0^x g(s)ds$. 
Since $g \geq 0$, $f$ is increasing (choose $g(x)> 0$ for $x> 0$ to be strictly increasing). Since $g(x) \rightarrow 0$ as $x\rightarrow 0^+$, $f^\prime(0)=0$ from the right. Since $\int g/M = 1$, $g$ will approach $0$ at infinity (otherwise the integral won't converge), so $f$ will approach $1$ and $f^\prime$ will approach $0$ at infinity.
To construct such a $g$ function, take $g = |B|\cdot I$, where $B$ is a continuous bounded function with $B(0)=0$ and $I$ is a continuous positive function such that $I(0) <\infty$ and $\int_0^\infty I < \infty$. Examples of $B$ functions include $\sin(x)$ and $\frac{\sqrt{x}}{\sqrt{x+1}}$, and examples of $I$ functions include $1/(x+1)^p$ for $p>1.$ 
