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

Question

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 $n \to \infty$?

Answer 1

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.

Answer 2

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.