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.

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    $\begingroup$ Are you asking in the contexts of standard calculus or also in a more advanced context (e.g. complex analysis)? $\endgroup$
    – user
    Apr 21, 2018 at 17:25
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    $\begingroup$ @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 $\endgroup$
    – Han
    Apr 21, 2018 at 17:30
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    $\begingroup$ 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. $\endgroup$
    – user
    Apr 21, 2018 at 17:49
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    $\begingroup$ @gimusi The question, as stated, is quite clear: a student, who has just learned about limits and derivatives, wants to know if a function can meaningfully be said to have a derivative at infinity. There is already a very good answer which addresses this question. Please don't ask the original poster to edit their question so that your answer more closely aligns with the question. It would be better to edit your answer so that it more accurately addresses the question. $\endgroup$
    – Xander Henderson
    Apr 21, 2018 at 17:52

1 Answer 1


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.

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    $\begingroup$ 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. $\endgroup$
    – user296602
    Apr 21, 2018 at 18:21
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    $\begingroup$ I don't get what one-point compactification is or Taylor series are, but aside from that, the answer is pretty understandable for me. $\endgroup$
    – Han
    Apr 21, 2018 at 19:27
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    $\begingroup$ @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. $\endgroup$
    – Xander Henderson
    Apr 21, 2018 at 19:31
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    $\begingroup$ 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 $y$? $\endgroup$
    – user856
    Apr 22, 2018 at 9:12
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    $\begingroup$ 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 $-\infty$. $\endgroup$ Apr 22, 2018 at 9:46

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