I was thinking about the exercise mentioned in the title:
Let $f \in C^{n}(\mathbb{R})$ for some $n \geq 0$ and further suppose that $f^{k}(0)=0$ for each $0 \leq k \leq n$ then show that $f^{(k)}(x)/|x|^{n-k} \to 0$ as $x \to 0$ for each $0 \leq k \leq n$
I understand that the condition $f^{(k)}(0)=0$ for all $0 \leq k \leq n$ gives us that the Taylor polynomial $P_{k}(x)$ is identically zero for $0 \leq k \leq n$. But I am having trouble estimating the error term $R_{k}(x)$, I believe that we have that
$$|R_{k}(x)| \leq \frac{M|x|^{k+1}}{(k+1)!}$$ but this doesn't help much. Is the statement of the exercise even true?
Consider the case $n=2$. Assume that $f \in C^2(\mathbf R)$, $f(0) = 0$, $f'(0) = 0$, and $f''(0) = 0$. Then $$ \lim_{x \to 0} f''(x) = 0$$ because $f''$ is continuous. In particular, for any $\epsilon > 0$ there exists a neighborhood $I$ of $0$ with the property that $x \in I$ implies $|f''(x)| < \epsilon$.
Let $\epsilon > 0$ be given and choose $I$ accordingly. Let $x \in I$. According to Taylor's theorem there exist points $\xi,\zeta \in I$ with the property that $$f(x) = \frac{f''(\xi)}{2}x^2,$$ and $$f'(x) = f''(\zeta) x.$$ Thus $$ x \in I\setminus \{0\} \implies |f(x)| \le \frac{\epsilon}{2}|x|^2 \implies \left| \frac{f(x)}{x^2} \right|< \epsilon$$ and $$ x \in I\setminus \{0\} \implies |f'(x)| \le \epsilon|x| \implies \left| \frac{f'(x)}{x } \right|< \epsilon.$$ This gives you the conclusion you need. You should be able to generalize this to arbitrary $n$ without too much difficulty.