Suppose we are given a continuos function $f(x)$ where $x \in [0,2]$, and the function $f(x)$ is $n$-th-order differentiable, for $n \in \mathbb{N}$ and $n>2$. Besides, we know that these derivatives satisfy that \begin{align} f(0) &= 0; \\ f'(0) &\neq 0;\\ f^{(i)}(0) &= 0, \text{ for } 2 \leq i \leq n. \end{align} Based on Taylor expansion, we might write \begin{align} f(x) &= \sum_{j=0}^{n-1} \frac{f^{(j)}(0)}{j!}(x-0)^j + \frac{f^{(n)}(\xi)}{n!}(\xi-0)^n \\ &= f'(0)x + \frac{f^{(n)}(\xi)}{n!}\xi^n, \end{align} where $0 < \xi < x$.
The question is that, according to the assumption of the derivatives of $f(x)$, can we infer that $f(x)=f'(0)x + O(x^n)$, or what can we speak about the error term $(f^{(n)}(\xi)/n!)\xi^n$ if we don't know whether $f^{(n)}(\xi)$ is bounded or not?
Yes you can. This is known as Taylor's theorem (http://en.wikipedia.org/wiki/Taylor%27s_theorem)
Basically, it says that if $f(x)$ is differentiable $n$ times, and let $$T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(x)}{k!}(x-x_0)^k $$ be the taylor polynomial, then
$$f(x) = T_n(x) + R_n(x)$$, where $R_n(x) \in O((x-x_0)^n)$
If $f(x)$ is differentiable another time, then you can write $$R_n(x) = \frac{f^{(n+1)}(\zeta)}{n+1!}(x-x_0)^{n+1}$$, with $\zeta \in [x_0, x)$
But still is $R_n \in O((x-x_0)^n)$