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Let me preface this by saying that 'convergence' refers to $\lim_{n\to\infty}\frac{d^nf(x)}{dx^n}$ being well defined. This stands in contrast to functions which, despite having derivatives of all orders, do not approach any particular value as each successive derivative is taken (e.g. $\cos{x}$ or $1/x$).

There are two trivial cases for the 'infinitieth' derivative - namely $ke^{x+c}$, for which $\frac{d^\infty f(x)}{dx^\infty}=ke^{x+c}$, and $\frac{d^\infty f(x)}{dx^\infty}=0$, which is the case for any function with some constant $n^\text{th}$ derivative (e.g. power/polynomial functions).

Intuitively, I can think of a few reasons why there wouldn't be any other examples, but then there might be some remarkable special function which defies all intuition.

Are there any non-trivial examples of functions (real or complex) where the 'infinitieth' derivative exists?

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    $\begingroup$ Presumably by "power functions" you mean "polynomials"? $\endgroup$ – Rob Arthan Feb 20 at 1:59
  • $\begingroup$ Polynomial functions, yes. $\endgroup$ – R. Burton Feb 21 at 0:07
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Assume that in some neighbourhood of $x=0$, the function $f$ is given by a convergent power series $f(x) = \sum_{n \in \Bbb N} a_n x^n$. Then the existence of $\lim_{n\to \infty} f^{(n)}(0)$ is equivalent to the existence of

$(*) \qquad \displaystyle\lim_{n\to \infty} n! \cdot a_n$.

Conversely, given a sequence $(a_n)_n$ such that $(*)$ exists, the function $f(x) := \sum_{n \in \Bbb N} a_n x^n$, by comparison with the exponential series, actually converges on all of $\Bbb R$, and the limit of its derivatives also exists everywhere.

The set of functions thus defined by sequences satisfying $(*)$ is closed under addition and scalar multiplication, and includes all examples so far (for polynomials, $a_n = 0$ for $n \gg 0$; for Robert Israel’s basic case $f(x) = Ae^{cx}$ with $c \in (-1, 1]$, we have $a_n = \frac{A c^n}{n!}$); but also many more. For example:

(1) $f(x) = \sum_{n \in \Bbb N} \frac{1}{(n!)^2} x^n$, or $f(x) = \sum_{n \in \Bbb N} \frac{1}{(n!)!} x^n$

(2) take any subset $S \subsetneq \Bbb N$ and set $f(x) = \sum_{n \in S} \frac{1}{n!} x^n$

or any combination of these ideas (Robert Israel's general example is included here too).

For the analogous question over $\Bbb C$, these should be all solutions. If you do not impose stronger conditions, on $\Bbb R$ there are more, e.g. $f(x) = e^{-1/x^2}$ (with the discontinuity at $0$ removed) at least has $\lim_{n\to \infty} f^{(n)}(0) = 0$.

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Also $f(x) = P(x) e^{cx}$ where $|c|<1$ and $P$ is a polynomial has $\lim_{n \to \infty} f^{(n)}(x) = 0$. And don't forget that linear combinations of solutions are solutions.

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