# Can you take the derivative of a function at infinity?

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

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.

• Are you asking in the contexts of standard calculus or also in a more advanced context (e.g. complex analysis)?
– user
Apr 21, 2018 at 17:25
• @gimusi standard calculus - I've only just learnt about limits and derivatives, and we haven't even touched on the subject of complex numbers in class; so nothing very advanced
– Han
Apr 21, 2018 at 17:30
• Thanks, maybe you should clarify that in your OP since many users didn’t understand that and are giving answers based on more advanced topics.
– user
Apr 21, 2018 at 17:49

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.

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.

• If the downvoter would share the reason for their vote, I'd greatly appreciate it. If there are any improvements to be suggested, I would be happy to hear them.
– user296602
Apr 21, 2018 at 18:21
• I don't get what one-point compactification is or Taylor series are, but aside from that, the answer is pretty understandable for me.
– Han
Apr 21, 2018 at 19:27
• @gimusi If a kindergartener asked "Can I subtract 5 from 3?", it would be very reasonable to answer "No. If you have three apples and try to give away five of them, you can't do that. Hence 5 can't be taken from 3." This would be an okay answer in the context of kindergarten mathematics. However, a better answer would be to point out that numbers can be generalized beyond cardinalities of sets, giving us negative numbers. This answer is the equivalent: it points out that the naive answer is "No," but that there is a very reasonable way of making things make sense. That answer is good. Apr 21, 2018 at 19:31
• As someone unfamiliar with analytic capacity, I find this answer completely unilluminating. There is a big gap between the first paragraph, which gives a quite reasonable interpretation of $f(\infty)$, and the second, where the formula for $f'(\infty)$ is simply asserted with no explanation. Sure, it makes sense that a derivative should have a term of the $f(x)-f(\infty)$ in it, but why is it multiplied by $x$? Why should we interpret it as differentiating $f(1/x)$ at $x=0$, when $f'(y)$ is not the same as differentiating $f(1/x)$ at $x=1/y$ for finite $y$?
– user856
Apr 22, 2018 at 9:12
• It's not clear why you would choose the one-point compactification over the two-points compactification $[-\infty, +\infty]$. I don't see why you would want to exclusively consider functions that exhibit the same behavior around $+\infty$ and $-\infty$. Apr 22, 2018 at 9:46