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

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.

## 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.

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 $$\text{radius} = \frac d {x + \frac 1 x - 2}.$$

• Well, just define something that works on $(0,1)$ and extend it by symmetry. $f(x)=\frac 1{1-x}-1$ for instance, but that choice isn't terribly important... all you need is the right behavior at $0$ and $1$.
– lulu
Jul 11, 2023 at 12:16
• @lulu: you beat me by 56 seconds! Jul 11, 2023 at 12:20
• Maybe something like ${1\over|\log(x)|}$?
– abc
Jul 11, 2023 at 12:30

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.

• Thanks. To meet the positive criteria, I can use $h(x) = [f(x)]^2$. Jul 11, 2023 at 12:27
• @SRobertJames There's no need to, because for $x>0$ we have $g(x) \geq 2$ and so $f(x)>0$ Jul 11, 2023 at 12:33
• Interesting: For at least one (unknown) value of $d$, the answer seems to be $[\frac {1+x}{1-x}]^2 - 1$, which, if my division is correct, equals $4 \cdot f(x)$. Is that correct? Jul 11, 2023 at 12:59
• This indeed seems to be the function! - See updated post. Jul 11, 2023 at 13:21

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.)

• Thank you. It's worth noting that $\frac x {(x-1)^2} = \frac 1 {x + \frac 1 x - 2}$. Jul 11, 2023 at 12:53
• This indeed seems to be the function! - See updated post. Jul 11, 2023 at 13:20

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 $$f(x) = \begin{cases} g(x), & \text{if x\in(0,1)} \\ g(1/x), & \text{if x\in(1,\infty)} \\ \end{cases}$$

• Yes, please see update that I'm trying to avoid piecewise. Jul 11, 2023 at 12:23