Are there non-trivial examples of an 'infinitieth' derivative? What is the criteria for 'convergence'?

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.

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 $\lim_{n\to \infty} f^{(n)}(0) = 0$.