Observation about convergent functions?

Question

I noticed this first in the case of the function $\arctan x + \arctan \frac{1}{x} = \frac{\pi}{2}$. This identity is true for all $x$, and it got me thinking about functions that are asymptomatic in general. Is the following statement true?

$\forall f(x) : \lim_{x\rightarrow \infty} f(x) = k\in \Bbb{R}, f(x) + f(\frac{1}{x}) = k$

I tested it with the function $f(x) = 1- 1/2^x$, and although it was very close, it was not exactly correct as it was in the case of $\arctan$. Why is this?

Answer 1

The statement is very restrictive and false in general. In fact, I suspect it defines a family of $\arctan$-like functions; with some additional hypotheses, it might even characterize $\arctan$! But for a wealth of counterexamples, try rational functions. If $f(x) = p(x)/q(x)$ where $p$ and $q$ are two polynomials of equal degree $n$ and leading terms $a$ and $b$, respectively, then $$\lim_{x\to\infty} f(x) = a/b$$ but $f(x) + f(1/x)$ may very well be anything it wants. For example, if $$f(x) = \frac{x^n+1}{x^n-1}$$ then $$f(x) + f(1/x) = 0$$ for all $x$, as it does for any choice of $p$ and $q$ such that $x^n p(1/x) = p(x)$ and $x^n q(1/x) = -q(x)$ (these are the "reverse" polynomials).

Your example $1 - 2^{-x}$ is also a counterexample, because $1 - 2^{-x} + 1 - 2^{-1/x} = 1$ iff $1 = 2^{-x} + 2^{-1/x}$, which is false since the RHS is non-constant.

But, in general, if $f(x) \to k$ (as $x \to \infty$) and if $f$ vanishes at the origin, then $$\lim_{x\to\infty} f(x) + f(1/x) = \lim_{x\to\infty}f(x) + \lim_{x\to 0^+} f(x) = k + 0 = k.$$ Since these hypotheses are quite lax, you can conclude that for "many" functions it is true that for all sufficiently large $x$, the expression $f(x) + f(1/x)$ will be very nearly equal to $k$.