Consider $f(x)=\exp(-\frac{1}{x^2})$ with $f(0)=0$. By induction one can verify $f$ is $C^\infty$, and $f^{(k)}(0)=0$ for all $k$. Therefore, $f$ cannot be approximated its Taylor expansion of any order around $0$.
Intuition:
$f(x)$ is very flat around $x=0$ and satisfies $\lim_{x\rightarrow 0}f(x)x^{-k} = 0$ for all $k$.
In the Taylor expansion, the remainder is $f$ itself.
Categories:
Analysis
Next:
Locally compactness