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Is it possible to have a formula that describes the following: $0\le y<1$ and $0\le x<\infty$ There is an asymptote at $y = 1$; so as $x$ approaches $\infty$, $y$ approaches $1$.

I have played with all manner of $y=1/x^n$ and $x=1/y^n$, square root combinations I can think of and I cannot get it to work.

Gary
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Bryon
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  • How about $ \ y \ = \ \frac{2}{\pi}·\arctan(x) \ \ $ on $ \ [0 \ , \ +\infty) \ \ ? $ –  Jan 20 '23 at 08:10

1 Answers1

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You can take$$\begin{array}{rccc}f\colon&[0,\infty)&\longrightarrow&\Bbb R\\&x&\mapsto&\dfrac x{x+1}.\end{array}$$Then the range of $f$ is $[0,1)$ and $\lim_{x\to\infty}f(x)=1$.

José Carlos Santos
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