Growth of Derivative Assume that $f \in C^1(\mathbb{R})$ such that $\lim_{x \rightarrow 0} x^{-(k+1)} f(x) = 0$ for some $k \in \mathbb{N} = \{1, 2, 3, \ldots \}$. Do we have $\lim_{x \rightarrow 0} x^{-k} f'(x) = 0$?

I suppose this question is really simple, but I'm struggeling with finding a counterexample or proof for this statement even if $k = 1$. Any suggestions?
I think I found the counterexample just after I posted the question: Take $f(x) := x^3 \sin(1/x)$. We have $f(x) / x^2 = x \sin(1/x) \rightarrow 0$ as $x \rightarrow 0$, but $f'(x) = 3x^2 \sin(1/x) - x \cos(1/x)$ and hence $f'(x) / x$ doesn't converge as $x \rightarrow 0$.
$lim_{x→0}\,\,x^{−(k+1)}f(x)=0$

take derivative with respect to $x$

$lim_{x→0}\,\,-(k+1)x^{−(k)}f(x) + x^{−(k+1)}f'(x)=0$

first part is zero for $k \in \{2,3,4...\}$

so 

$lim_{x→0}\,\, x^{−(k+1)}f'(x)=0 \,\, \forall k \in \{2,3,4...\}  $

thus we have 

$lim_{x→0}\,\, x^{−k}f'(x)=0 \,\, \forall k \in \{3,4,5...\}  $