Every derivative of $e^{(-1/x^2)}$ at the origin is zero. Yet, this is not a constant function. Clearly, the limit as $x$ diverges to either side is $1,$ yet the function at $0$ is $0$ by continuous extension.
My intuition is that if every derivative is zero, then the function can never change. Why is my intuition wrong?
Let's begin with a precise statement of Taylor's theorem: if $f$ is $k$ times differentiable at $a\in\Bbb R$, some function $h_k$ satisfies$$f(x)=\sum_{j=0}^k\frac{f^{(j)}(a)}{j!}(x-a)^j+h_k(x)(x-a)^k,\,\lim_{x\to a}h_k(x)=0.$$With $a:=0$ and $f:=\exp\frac{-1}{x^2}$, defined at $0$ by continuity viz. $f(0)=0$, the polynomial is identically $0$, and $h_k(x)=x^{-k}f(x)$ for any integer $k\ge0$. So $f$ doesn't have to be identically $0$; it just has to decay faster than any polynomial as you approach $0$.