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I want to design an increasing monotonic function asymptotic to $1$ when $x\to +\infty $ that uses a logarithm.

Also, the function should have "similar properties" to $\dfrac{x}{1+x}$, i.e.:

  1. increasing monotonic
  2. $f(x)>0$ when $x>0$
  3. gets close to 1 "quickly", say $f(10)>0.8$
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Does $1-\frac1{\log\,x}$ not do what you want? –  J. M. Apr 8 '13 at 13:09
    
Uhm, it would be actually great to have a function that "imitates" $\dfrac{x}{1+x}$, in the sense of having also these two additional properties: (1) $f(x) > 0$ when $x > 0$ and (2) which gets close to 1 quickly, say $f(10) > 0.8$. Maybe I should edit the question? –  Alphaaa Apr 8 '13 at 13:26
    
How about $f(x)=1-(\log x/x)$? –  Gerry Myerson Apr 8 '13 at 13:26
1  
Yes, you should have included those pieces of information in your question to begin with... –  J. M. Apr 8 '13 at 13:30
1  
In $\frac{x}{x+1}$. replace $x$ by some high enough power of $\log(x+1)$ to get the value at $10$ that you desire. –  André Nicolas Apr 8 '13 at 13:43
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1 Answer 1

up vote 2 down vote accepted

How about $$ f(x)=\frac{a\log(1+x)}{1+a\log(1+x)},\quad a>0? $$ Notes:

  1. I have used $\log(1+x)$ instead of $\log x$ to avoid issues near $x=0$ and to make it more similar to $x/(1+x)$.
  2. Choose $a$ large enough to have $f(10)>0,8$.
  3. You can see the graph of $f$ (for $a=1$) compared to $x/(1+x)$.

enter image description here

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