# Does $\lim_{n \to \infty}\frac{\mathrm d^n}{\mathrm dx^n} f(x)$ converge?

Is there a general way to show whether the limit $$\lim_{n \to \infty}\frac{\mathrm d^n}{\mathrm dx^n} f(x)$$ converges to some expression?

What about repeatedly integrating an expression $n$ times as $n \to \infty$?

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For the notion of limit, you need a notion of convergence or of "distance." Here, you are talking about functions, not numbers, so you need to specify how you measure the "distance" between two functions. Turns out, there are many different ways of doing it, and the answer for a particular $f(x)$ may depend on which way you do it. So you would need to specify that in general. Among the different kinds of "convergence" you can read about there is uniform convergence, pointwise convergence, convergence in measure, and convergence in $\mathcal{L}_p$-norm. But this hardly exhausts the possibilities.

Once you have the notion of convergence of functions, then you are dealing with a sequence of functions, and one checks convergence the "usual" way (that is, in essentially the same ways as you do with sequences of real numbers, substituting the absolute value of the difference with whatever way you have decided to describe distance of functions).

As for "repeated integrals", that's not going to help much. The $n$th iterated integral of $\frac{d^n}{dx^n}f(x)$ is $f(x)+p(x)$, where $p(x)$ is an arbitrary polynomial of degree $n-1$ (which you get through the constants of integration). Taking "the limit" as $n$ goes to infinity just drops you back into the problem of "what does it mean for a sequence of functions to converge"? Plus, an "infinite polynomial" (a power series) is probably not what you want anyway.

Now, for specific functions one may be able to discern either a pattern, or an answer. For polynomials, it is plain that $f^{(n)}(x)$ will be zero for all sufficiently large $n$, and in that case no matter how you define your distance between functions, the answer will be the zero function. If you have a function with periodic derivatives, such as $\sin(x)$, you know there will be no limit. And so on.

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Thank you for a thorough explanation! It appears the question wasn't quite as basic as I thought. Would I learn more about this in an introductory Real Analysis course? (I am a physics student who has so far only taken Calculus and am taking Linear Algebra and Abstract Algebra next semester.) –  Aqwis Dec 9 '10 at 17:06
@Aqwis: I was actually going to add to the answer, thinking belatedly that perhaps I had misinterpreted the question and you were asking for ways to figure out a formula for the "general term" that occurs in a Taylor series; glad that wasn't the case. Real Analysis will probably cover things like pointwise and uniform convergence, maybe even a bit of $\mathcal{L}_p$-convergence. Convergence in measure is covered in Measure Theory/Functional Analysis (both of which will ask you for Real Analysis as a prerequisite). But yes, you would learn more there. –  Arturo Magidin Dec 9 '10 at 18:22

In terms of pointwise convergence, you cannot ensure the limit to exist at a given point, even if all the derivatives of $f$ exist (everywhere).

For example, by a theorem of Borel, any sequence $a_0,a_1,\dots$ is $f(0),f'(0),f''(0),\dots$ for some function $f$, see this question.

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On the other hand, I think if d^n/dx^n f(x) converges pointwise at x = 0, then it must in fact converge uniformly on compact sets. Possibly it converges uniformly... –  Qiaochu Yuan Dec 10 '10 at 5:12
@Qiaochu: Do you mean converges pointwise on an interval containing x = 0? If not, I'm not so sure about uniform convergence on compact sets. For instance, what about non-analytic smooth functions? –  Jesse Madnick Dec 10 '10 at 7:39
Hmm, on the other hand, {0} is a compact set.... –  Jesse Madnick Dec 10 '10 at 7:41
@Jesse: sorry, I forgot to add hypotheses. I'm thinking about the case that f is real analytic with infinite radius of convergence, which is maybe too special a case... –  Qiaochu Yuan Dec 10 '10 at 7:49
@Qiaochu: Yes, in that case things are significantly different/easier. –  Andres Caicedo Dec 10 '10 at 15:09