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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?

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    $\begingroup$ Do you mean logistic functions? Or maybe you mean the cumulative distribution function for a (real-valued) random variable $X$? $\endgroup$ – Hirshy Jul 6 '15 at 8:23
  • $\begingroup$ The function must be bounded, like $3-e^{-x}<3$ or $\frac{\ln(x)}{\ln(x+1)}<1$ or $\arctan(x)<\pi/2$. $\endgroup$ – Yves Daoust Jul 6 '15 at 8:24
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    $\begingroup$ $y=a \tan^{-1}(bx+c)+d$, $y=a \tanh(bx+c)+d$ may be $\endgroup$ – Claude Leibovici Jul 6 '15 at 8:27
  • $\begingroup$ @travis: is your example-function growing ? And if I am right, $f(x)<1$ for all $x>0$. $\endgroup$ – Yves Daoust Jul 6 '15 at 8:41
  • $\begingroup$ Ah, you're right, I didn't parse "growing" as "increasing", cheers. $\endgroup$ – Travis Willse Jul 6 '15 at 8:42
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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 :

$$C-\frac{1}{x^2+1}$$ $$C-e^{-3x}$$

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  • $\begingroup$ But didn't OP ask for increasing functions? $\endgroup$ – coldnumber Jul 6 '15 at 8:29
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    $\begingroup$ @ColdNumber that's why I say "substract" ! $\endgroup$ – Xoff Jul 6 '15 at 8:33
  • $\begingroup$ Whoops! you're right. $\endgroup$ – coldnumber Jul 6 '15 at 8:34
  • $\begingroup$ @Xoff Better say: subtract. $\endgroup$ – CiaPan Jul 6 '15 at 8:39
  • $\begingroup$ @CiaPan Yes, sorry, I'm not so good in english sometimes $\endgroup$ – Xoff Jul 6 '15 at 8:45
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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 $\lim\limits_{x\to\infty}f(x)=1$.

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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 $a>b \implies f(a)>f(b)$.

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  • $\begingroup$ But this function is only monotonic on either $\mathbb R^*_-$ or $\mathbb R^*_+$, so only on restricted domains. $\endgroup$ – Hirshy Jul 6 '15 at 8:30
  • $\begingroup$ Yes, thanks, I'm editing my answer. $\endgroup$ – coldnumber Jul 6 '15 at 8:30
  • $\begingroup$ @pjs36 for the $\ln(x)$ example the domain is $D=\mathbb R_+^*$ and $f$ is monotonic on its (full) domain. This is not the case with $f(x)=\frac{1}{x}$ which IMO is a difference. $\endgroup$ – Hirshy Jul 6 '15 at 8:48
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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 $x\to\infty$.

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