I was just looking for a sequence $\{f_{n}\}\subset C_{c}^{\infty}(\mathbb{R})$ of smooth functions of commpact support on $R$ and a smooth $f\in C^\infty(\mathbb{R})-C_{c}^{\infty}(\mathbb{R})$ of not compact support such that for all $k\in\mathbb{N}$ $\sup_{x\in \mathbb{R}}\{|f_{n}^{(k)}(x)-f^{(k)}(x)|\rightarrow 0$.
I took a function $$f_{n}(x)=\begin{cases} -1 & -1\leq x\leq -\frac{1}{n} \\ nx & -\frac{1}{n}\leq x\leq \frac{1}{n}\\ 1 & \frac{1}{n}\leq x\leq 1 \\ 0 &|x|>1 \end{cases} $$ and $$f(x)=\begin{cases} 0 & -1\leq x\leq 1 \\ 1 & |x|>1 \\ \end{cases} $$ where $f_{n}$ have compact support $[-1,1]$ and $f$ have no compact support and meeting the requirement. Is my example correct. I'm not so sure on the smoothness condition stated. Thank you
Hint for a start: Suppose $g \in C(\mathbb R),$ with $g=1$ on $[-1,1],$ and $g(x) = 0$ for $|x|\ge 2.$ Let $f\in C_0(\mathbb R).$ Can you show $[1-g(x/n)]f(x) \to 0$ uniformly on $\mathbb R?$