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 $\lim _{x\to \infty \:}f\left(x\right)=\mathrm{constant}$

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

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 :

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

• But didn't OP ask for increasing functions? – coldnumber Jul 6 '15 at 8:29
• @ColdNumber that's why I say "substract" ! – Xoff Jul 6 '15 at 8:33
• Whoops! you're right. – coldnumber Jul 6 '15 at 8:34
• @Xoff Better say: subtract. – CiaPan Jul 6 '15 at 8:39
• @CiaPan Yes, sorry, I'm not so good in english sometimes – Xoff Jul 6 '15 at 8:45

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 $\lim\limits_{x\to\infty}f(x)=1$.

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

• But this function is only monotonic on either $\mathbb R^*_-$ or $\mathbb R^*_+$, so only on restricted domains. – Hirshy Jul 6 '15 at 8:30
• Yes, thanks, I'm editing my answer. – coldnumber Jul 6 '15 at 8:30
• @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. – Hirshy Jul 6 '15 at 8:48

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 $x\to\infty$.