Let $f:D \to \mathbb{R}$ be a real function with $U \subset D$ an open interval. Suppose that $f$ is $k$ times differentiable at $a \in U$. Then
$$R_k(x) \in o(|x-a|^k) \ \text{as} \ x \to a$$
where $R_k(x)=f(x)-P_k(x)$ and $P_k$ is the $k$th order Taylor polynomial of $f$ evaluated at $a$.
Like many, I am somewhat averse to the usage of L'Hospital's rule. Is there a proof of the above without it? Note the minimal conditions on $f$.
Let's do it with $k=3$ and $a=0$ to illustrate. To simplify, let $g(x)=R_3(x).$ Then $0 = g(0)= g'(0)= g''(0)=g'''(0).$ By the MVT,
$$g(x)= g(x)-g(0) = g'(c_1)x = (g'(c_1)-g'(0))x$$ $$ = g''(c_2)c_1x = (g''(c_2)-g''(0))c_1x = \frac{g''(c_2)-g''(0)}{c_2}c_2c_1x.$$
As $x\to 0,$ $c_1,c_2$ are dragged along to $0$ with it. We have $c_2c_1x = O(x^3)$ and the fraction on the right $\to g'''(0) = 0.$ Thus $g(x) = o(x^3)$ as desired.