What function $f(x) = f(1/x)$ has limits $\lim_{x \to 0} f(x) = 0, \lim_{x \to \infty} f(x) = 0$, and $\lim_{x \to 1} f(x) = \infty$?

Question

What (simple) function has the following properties:

• It maps the positive reals (except for 1) to positive reals: $f: (0,1) \cup (1, +\infty) \to (0, +\infty)$.

• It is symmetric about reciprocals: $f(x) = f(1/x)$.

• It has limits $\lim_{x \to 0} f(x) = 0, \lim_{x \to \infty} f(x) = 0$, and $\lim_{x \to 1} f(x) = \infty$.

• It may incorporate one arbitrary constant (a parameter), which we'll call $d$, ranging from $(0, + \infty)$. The function $f$ is linear in $d$.

Surprisingly, I'm not able to construct any function which has all of those properties.

Background: In the context of a particular problem, I've been able to show that the solution involves a function $f$. I've yet to figure out what $f$ is, but I can show it has the properties above. I'd like to use them to form a candidate function, which I can then perhaps use as an ansatz or refine to get the solution.

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Clarification

Of course, I could piecewise construct such a function, but this is unlikely to be a good ansatz. I'm looking for a function which has these properties "naturally", for some definition of "naturally". As an inspiration, consider the gamma function: There are many (infinite) functions that equal factorial on the integers, but only one "natural" function.

I don't know how to define "natural" - if I did, that would probably be my answer - but piecewise is certainly not.

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More information

For those interested, the problem I'm working on is the Circle of Apollonius (Interactive), which has complex equation $$|z - z_0| = \rho |z - z_1|.$$

I was trying to find a function that maps the constant $\rho$ to the radius of the circle, and deduced that it must have the above properties (with $d = |z_1 - z_0|$).

Indeed, it seems the answers from "Fluffy Alpaca" and aschepler are correct, in that the radius seems to be proportional to $$\frac x {(x-1)^2} = \frac 1 {x + \frac 1 x - 2}$$ where $x = \rho^2$. I've yet to determine the constant of proportionality, but assume it is linear in $d$ (because the radius must be linear in some measure of distance, which can only come from $d$), suggesting $$\text{radius} = \frac d {x + \frac 1 x - 2}.$$

Answer 1

Let $$g(x)=x+\frac{1}{x}$$ $$f(x)=\frac{1}{g(x)-2}$$

This should work - try playing around a bit with it on desmos for example.

Answer 2

Take any function $g:(0,1)\to(0,\infty)$ that has $\lim_{x\to 0}g(x)=0$ and $\lim_{x\to 1}g(x)=\infty$. For instance, $g(x)=dx/(1-x)$ with $d\in(0,\infty)$. Then just define $$f(x) = \begin{cases} g(x), & \text{if $x\in(0,1)$} \\ g(1/x), & \text{if $x\in(1,\infty)$} \\ \end{cases}$$

Answer 3

One function satisfying these properties, and without "definition by cases", is

$$ f(x) = \frac{d}{(\ln x)^2} $$

This comes from noticing $\lim_{x \to 0^+} \ln x = -\infty$, $\ln 1 = 0$, and $\lim_{x \to +\infty} \ln x = +\infty$, then applying the power $-2$ to get the desired limits.

Or along the same lines,

$$ f(x) = \frac{dx}{(x-1)^2} $$

(That's $d$ times $x$, not a differential form.)