What kind of growing function has a constant as limit? My knowledge in mathematics are a bit old and I'm looking for functions with constant as limit. The function must always grow.
The curve should be something similar to $\sqrt{x}$ or $\ln(x)$ but with $\lim _{x\to \infty \:}f\left(x\right)=\mathrm{constant}$
Could you please help to find this kind of functions?
 A: If you just want to have some functions with the asked properties, you can easily construct one as Xoff pointed out. If you are looking for a class of functions (possibly from applied mathematics), the logistic function $$f:\mathbb R\rightarrow\mathbb R,~f(x)=\frac{L}{1+e^{-k(x-x_0)}}$$ might be something you are looking for (for the meaning of the constants $L,k,x_0$ consider the wikipedia page, it has lots of information on this). 
If your familiar with stochastics and random variables, you could take a look at the cumulative distribution function. It is monotonicically non-decreasing with $\lim\limits_{x\to\infty}f(x)=1$.
A: An example could be $f(x)=-1/x$. It grows toward 0, its asymptote.
EDIT: As Hirshy pointed out, $f$ is not an increasing function if its domain is $\mathbb{R}\setminus\{0\}$. 
So, define $f : \mathbb{R}_+ \to \mathbb{R}$ by $f(x)=-1/x$. 
This function increases monotonically: If $a>b>0$, then $1/a<1/b$, and $-1/a>-1/b$, which means that $a>b \implies f(a)>f(b)$. 
A: The simplest way is to consider any decreasing function such that its limit is $0$, like $\frac{1}{x+1}$ or $e^{-x}$, and subtract from a constant :
$$C-\frac{1}{x^2+1}$$
$$C-e^{-3x}$$
A: Functions of $x$: $$-e^{-x}$$ $$ -x^{-1}$$ $$\cos \tfrac 1x \ \ \text{for}\ x > \tfrac 1\pi $$ $$\arctan x$$ all are monotonically growing and have finite limit as $x\to\infty$.
