n'th derivative does not vanish, but $\lim_{n\to \infty} f^{(n)}=0$. Let $f\,$$\in$$\,C^\infty[\mathbb{R},\mathbb{R}]$ .  Apparently the only functions $f$ for which there exists $n\in\mathbb{N}$ such that $f^{(n)}=0$ are polynomials in $\mathbb{R}[x]$. 
Is it possible to characterize the functions $f\,$$\in$$\,C^\infty[\mathbb{R},\mathbb{R}]$ for which $\lim_{n\to \infty} f^{(n)}=0$, but $f^{(n)}\neq 0$ for all $n\in\mathbb{N}$. For example are they dense in ($C^\infty[\mathbb{R},\mathbb{R}]$,$||.||_{\infty}$)?
EDIT: It maybe easier if we resctrict to $C^\infty[(0;1),\mathbb{R}]$. Any discussion is welcome for this one too.
 A: First,
a few remarks:


*

*$(C^\infty(\mathbb R,\mathbb R),\|\cdot\|_\infty)$ is problematic:
there are many smooth functions from $\mathbb R$ to itself that are unbounded,
such as $f(x)=x$.
Thus,
$\|\cdot\|_\infty$ does not define a norm on $C^\infty(\mathbb R,\mathbb R)$.

*The same problem arises on $(C^\infty((0,1),\mathbb R),\|\cdot\|_\infty)$,
where $(0,1)$ denotes the open interval.
Consider $f(x)=1/x$.

*$\|\cdot\|_\infty$ does indeed define a norm on $C^\infty([0,1],\mathbb R)$,
where $[0,1]$ denotes the closed (compact) interval.
However,
it makes no sense to ask if a subset $A$ is dense in $(C^\infty([0,1],\mathbb R),\|\cdot\|_\infty)$:
Since we're considering the norm topology induced by $\|\cdot\|_\infty$
(which is in fact a metric topology with $d(x,y)=\|x-y\|_\infty$),
the set $A$ is dense in $C^\infty([0,1],\mathbb R)$ if and only if $\overline A=C^\infty([0,1],\mathbb R)$ (where $\overline A$ denotes the closure).
However,
this can never be satisfied since $C^\infty([0,1],\mathbb R)$ is not itself closed under $\|\cdot\|_{\infty}$.
Indeed,
according to the Weierstrass Approximation theorem,
one can find a sequence of polynomial functions $\{p_n\}\subset C^\infty([0,1],\mathbb R)$ such that $\|p_n-f(x)\|_{\infty}\to0$,
where $f(x)=|x-1/2|$,
yet $f$ is not differentiable.


An alternative question in the same spirit as what you're asking would be the following:
Question. What is the closure of the set $A$ of functions $f\in C^{\infty}([0,1],\mathbb R)$ such that
\begin{align}f^{(n)}\neq0 \text{ for every $n$ and }\lim_{n\to\infty}\|f\|_\infty=0.\tag{1}\end{align}
The answer to this question was provided in a comment by Willie Wong:
The set of functions $f:[0,1]\to\mathbb R$ defined as
$$f(x)=p(x)+\epsilon\sin(\epsilon x)$$
(where $p$ is a polynomial function and $0<\epsilon<1$)
is a subset of the functions that satisfy conditions $(1)$,
and it approximates the polynomial functions,
which are dense in $C[0,1]$.
Thus,
the closure of your set is the continuous functions on the compact [0,1].
Note. If you ask the same question as above but with the condition
\begin{align}f^{(n)}\neq0 \text{ for every $n$ and }\lim_{n\to\infty}f(x)=0 \text{ pointwise},\tag{2}\end{align}
then the result still holds,
since uniform convergence implies pointwise convergence,
hence the set of every function satisfying $(2)$ contains the set of every function satisfying $(1)$.
