Assume that $f'(x) \in C[0,1]$, $f^{(n)}(0)=0(\forall n\geq 0)$, and $\exists C>0,\forall x \in [0,1]:|xf'(x)|\leq C|f(x)|.$ Problem
Assume that $f'(x) \in C[0,1]$, $f^{(n)}(0)=0(\forall n\geq 0)$, and 
$$\exists C>0,\forall x \in [0,1]:|xf'(x)|\leq C|f(x)|.$$
Prove
 1. $\lim\limits_{x \to 0+}\dfrac{f(x)}{x^n}=0(\forall n \geq 0);$
 2. $f(x)\equiv 0(\forall x \in [0,1]).$
Attempt
The first one is too easy, if we apply L'Hôpital's rule repeatedly or apply Taylor's formula. Hence, we only need to focus on the second.
Since $f'(x) \in C[0,1]$ and $f(0)=0$，By Lagrange's theorem，$$\forall x_0 \in (0,1]:|f(x_0)|=|f(0)+f'(\xi_1)(x_0-0)|=|x_0f'(\xi_1)|.$$
Therefore
$$|x_0f'(x_0)|\leq C|f(x_0)|=C|x_0f'(\xi_1)|,$$
which implies
$$|f'(x_0)|\leq C|f'(\xi_1)|,0<\xi_1<x_0\leq 1.$$
Repeat the process above. We obtain
$$|f'(x_0)|\leq C|f'(\xi_1)|\leq C^2|f'(\xi_2)|\leq \cdots \leq C^n|f'(\xi_n)|,$$
where $$0<\xi_n<\cdots<\xi_2<\xi_1<x_0\leq 0.$$
If $C<1$, notice that $f'(\xi_n)$ is bounded,then 
$$|f'(x_0)|=\lim_{n \to \infty}|f'(x_0)|\leq \lim_{n \to \infty}C^n|f'(\xi_n)|=0,$$
which implies $$\forall x_0 \in (0,1]:f'(x_0)=0.$$But $f(0)=0$,hence
$$\forall x \in [0,1]:f(x)\equiv 0.$$
How to go on with this? What if $C \geq 1$? 
 A: Once you have part 1, this is how part 2 can be shown:
Assume $f(a)\ne 0$ for some $a\in(0,1)$. Then $f(x)$ has constant non-zero sign on an interval $(u,v)\ni a$ and  $h(x)=\ln |f(x)|$ is in $C^1(u,v)$ (namely, with derivatvie $h'(x)=\frac{f'(x)}{f(x)}$). 
By taking the smallest possible $u$, we can assume that $f(u)=0$ and hence $$\tag1\lim_{x\to u^+}h(x)=-\infty.$$
If $u>0$, we have $|h'(x)|=\frac{|f'(x)|}{|f(x)|}\le \frac Cx<\frac Cu$ on $(u,v)$, so that $h'$ and then also $h$ is bounded on $(u,v)$, contradicting $(1)$. We conclude that $u=0$.
For $0<x<v$, let $g(x)=C\ln x$.
Then for $0<x\le a$, we have
$$ h'(x)=\frac{f'(x)}{f(x)}\le \frac Cx=g'(x)$$
and conclude $h(x)-h(a)\ge g(x)-g(a)$ for $0<x\le a$.
Therefore $$|f(x)|=e^{h(x)}\ge e^{g(x)-g(a)+h(a)}=\underbrace{e^{h(a)-g(a)}}_{>0}\cdot x^C\qquad\text{for }0<x\le a $$
If $n>C$, this implies that 
$$0=\lim_{x\to0^+}\frac{|f(x)|}{x^n}=e^{h(a)-g(a)}\cdot \lim_{x\to0^+}x^{C-n}= +\infty ,$$
contradiction. Therefore, $a$ with $f(a)\ne 0$ does not exist.
