calculus - Can you take the derivative of a function at infinity?

Exactly the title: can you take the derivative of a function at infinity?

I asked my maths teacher, and while she thought it was an original question, she didn't know the answer, and I couldn't find anything online about this.

Maybe this is just me completely misunderstanding derivatives and functions at infinity, but to me, a high schooler, it makes sense that you can. For example, I'd imagine that a function with a horizontal asymptote would have a derivative of zero at infinity.

Answer

In a very natural sense, you can! If $\lim_{x \to \infty} f(x) = \lim_{x \to -\infty} f(x) = L$ is some real number, then it makes sense to define $f(\infty) = L$, where we identify $\infty$ and $-\infty$ in something called the one-point compactification of the real numbers (making it look like a circle).

In that case, $f'(\infty)$ can be defined as

$$f'(\infty) = \lim_{x \to \infty} x \big(f(x) - f(\infty)\big).$$
When you learn something about analytic functions and Taylor series, it will be helpful to notice that this is the same as differentiating $f(1/x)$ at zero.

Notice that this is actually not the same as $\lim_{x \to \infty} f'(x)$.

These ideas actually show up quite a bit in analytic capacity, so this is a rather nice idea to have.

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I wanted to expand this answer a bit to give some explanation about why this is the "correct" generalization of differentiation at infinity. and hopefully address some points raised in the comments.

Although $\lim_{x \to \infty} f'(x)$ might feel like the natural object to study, it is quite badly behaved. There are functions which decay very quickly to zero and have horizontal asymptotes, but where $f'$ is unbounded as we tend to infinity; consider something like $\sin(x^a) / x^b$ for various $a, b$. Furthermore, $\lim_{x \to \infty} f'(x) = 0$ is not sufficient to guarantee a horizontal asymptote, as $\sqrt{x}$ shows.

So why should we consider the definition I proposed above? Consider the natural change of variables interchanging zero and infinity*, swapping $x$ and $1/x$. Then if $g(x) := f(1/x)$ we have the relationship

$$\lim_{x \to 0} \frac{g(x) - g(0)}{x} = \lim_{x \to \infty} x \big(f(x) - f(\infty)\big).$$

That is to say, $g'(0) = f'(\infty)$. Now via this change of variables, neighborhoods of zero for $g$ correspond to neighborhoods of $\infty$ for $f$. So if we think of the derivative as a measure of local variation, we now have something that actually plays the correct role.

Finally, we can see from this that this definition of $f'(\infty)$ gives the coefficient $a_1$ in the Laurent series $\sum_{i \ge 0} a_i x^{-i}$ of $f$. Again, this corresponds to our idea of what the derivative really is.

* This is one of the reasons why I used the one-point compactification above. Otherwise, everything that follows must be a one-sided limit or a one-sided derivative.