Which real functions have their higher derivatives tending pointwise to zero?

Let $\mathrm C^\infty\!(\Bbb R)$ be the space of infinitely differentiable functions $f:\Bbb R\rightarrow\Bbb R$, and define the subspace$$A:=\{f\in\mathrm C^\infty\!(\Bbb R):(\forall x\in \Bbb R)\lim_{n\rightarrow\infty} f^{(n)}(x)=0\},$$where $f^{(n)}$ is the $n$th derivative of $f\;(n=0,1,\dots).$ Clearly all polynomial functions are in $A$. Are any others?

Edit: Alfonso has answered this question well, but is there any characterization of $A$ in terms of familiar types of function?

Consider an analytic function $$\tag0f(x)=\sum_{n=0}^\infty \frac{a_n}{n!}x^n.$$ As $f^{(n)}(0)=a_n$, a necessary condition for such $f$ to be an example is that $a_n\to 0$.

But $a_n\to 0$ is also sufficient. Indeed, with $b_n:=\sup_{k\ge n} |a_k|$ we have $b_n\to 0$ and hence $$\begin{align}|f^{(n)}(x)|&= \left|\sum_{k=0}^\infty \frac{a_{n+k}}{k!}x^k\right|\\ &\le\sum_{k=0}^\infty\frac{|a_{n+k}|}{k!}|x|^k\\ &\le b_n\sum_{k=0}^\infty\frac{1}{k!}|x|^k\\&=b_ne^{|x|}\to 0. \end{align} $$ (Note that the calculation for $f^{(0)}(x)$ shows that $f$ is entire to begin with).

$$f(x)=e^{x/2}$$ $$f^{(n)}(x)=2^{-n}e^{x/2}$$