When does the $f^{(n)}$ converge to a limit function as $n\to\infty$? (This was something someone (almost) asked in a comment in a thread about repeated differentiation of polynomials.)
Consider a general smooth (that is $C^\infty$) function $f$. As usual $f^{(n)}$ denotes the $n$th derivative of $f$.
Under what circumstance does $\{ f^{(n)} \}_{n=1}^\infty$ converge to a limit function as $n$ goes to infinity?
For example when $0<k<1$ is fixed and $f(x)=e^{kx}$, then we have the pointwise convergence $f^{(n)} \to 0$ for $n\to\infty$ where $0$ is the zero function. The case $k=1$ is different.
Of course when $f(x)=\cos x$, there is no convergence of the $f^{(n)}$. But $\cos(kx)$ ...
Can a criterion be given?
 A: Partial answer
Let $f$ a real $C^\infty$ function s.t. $f$ can be expressed as its Maclaurin series with radius of convergence $R$ :
$$f(x)=\sum_{k=0}^{\infty}\frac {f^{(k)}(0)}{k!}x^k, \forall x \in (-R,R).$$
By properties of power series, in the interior of the domain of convergence, we have :
$$f^{(n)}(x)=\sum_{k=n}^{\infty}\frac {f^{(k)}(0)}{(k-n)!}x^{k-n}=\sum_{k=0}^{\infty}\frac {f^{(n+k)}(0)}{k!}x^{k} , \forall x \in (-R,R).$$
We claim the following :

If : $$\lim_{n \to \infty} f^{(n)}(0)=l ,l\in \mathbb{R}\qquad (*)$$ 
  Then $(f^{(n)})$ converges uniformly to $g(x)=l e^x$ on $(-r,r)$ for any $r$ s.t. $0<r<R$.

Proof
$$|f^{(n)}(x)-le^x|=\left| \sum_{k=0}^{\infty}\frac {f^{(n+k)}(0)}{k!}x^{k}-\sum_{k=0}^{\infty}\frac {l}{k!}x^{k}\right|$$
Then :
$$|f^{(n)}(x)-le^x| \leq \sum_{k=0}^{\infty}\left|\frac {f^{(n+k)}(0)-l}{k!}\right|r^{k}.$$
Since we have $(*)$, let $\varepsilon>0$, for $n$ big enough :
$$|f^{(n)}(x)-le^x|\leq \varepsilon e^r.$$
Since $x$ is arbitrary, the claim follows.
Remark: 
I believe nothing more can be done with less constraints on the sequence $(f^{(n)}(0))$. For example, if you suppose it only bounded, then consider $f(x)=\cos(x)$.
