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The expected curve looks like below.
^                   
|2____________________ 
|                                      o              x      
|                 o           x            
|       o       x               
|   o  x                 
| ox                  
|1____________________ 
|                   
|                   
|                   
|                   
|
_____________________> 

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Do you require that $f(0)=1$? –  fidbc Feb 24 '13 at 1:05
    
Yes, it's better be F(0)=1. But, compared to having the curve less inward-bended, F(0)=1 is not a must. –  Rick Chin Feb 24 '13 at 2:27

2 Answers 2

Take a function which you already know has a similar behaviour and then try to adapt it. For example take $g(x)=-\frac{1}{x}$, we know $\lim_{x\rightarrow\infty} g(x)=0$. It is now just a matter of "shifting this upwards and to the left", so that $\lim_{x\rightarrow\infty} f(x)=2$ and $f(0)=1$. For this example you can try using $f(x)=-\frac{1}{x+1}+2$ and see that this function has the desired property.

Note that here $f(x)=g(x+1)+2$, where $g(x+1)$ shifts the graph of $g$ one unit to the left and the $+2$ shifts it 2 units upwards.

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Thanks for guiding. It works. But, it would be ideal to have a less inward-bended curve, so Y value steps up more obvious when getting close to Y=2. If we allow the curve to hit and go through Y=2, which would only happen in a too large X value, is there a chance to make a less inward-bended curve? Thanks. –  Rick Chin Feb 24 '13 at 2:24
    
You have to be a bit more precise about what you mean by "inward-bended". One way of bending the curve would be by taking $g(x)=-\frac{k}{x}$ for some constant $k$. Depending on the type of "bending" you want you should choose an appropriate $k$. For this function, if you shift it up by $\epsilon>0$ then it will take the value 2 for some $x$, but it will take values greater than 2 afterwards since it is an increasing function on the non negative real numbers. –  fidbc Feb 24 '13 at 4:19

Another function is $y=1+(2/\pi)\arctan x$. Another is $y=2-e^{-x}$. Another is $$y={2\root3\of x+1\over\root3\of x+1}$$ If you can translate the phrase "inward-bended" into something precise and mathematical, maybe we'd come up with an answer better suited to your needs.

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Thanks. I added an additional "o" curve into above graph. The o curve is more inward-bended than the x curve. I guess g(x)=-k/x is less inward bended than f(x)=-1/x. What I wanted was something similar to the x curve. So, as X value increases, the Y value would not step up so steeply and can slowly approach an unreachable limit of Y=2. –  Rick Chin Feb 25 '13 at 5:14
    
So, the three examples I suggested --- how do they look to you? –  Gerry Myerson Feb 25 '13 at 5:31

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