infinite limit cycles 
Can we find differential equations in the real plane and class $C^{k}(\mathbb{R})\ k\geq 1 $" that have an infinity of limit cycles accumulating in the origin with the origin as singular point?

I think is true, my idea was, I know that
$$ \dot{x_{1}}=x_{2}+x_{1}(1-x_{1}^{2}-x_{2}^{2})\\ 
\dot{x_{2}}=-x_{1}+x_{2}(1-x_{1}^{2}-x_{2}^{2})$$
have a cycle limit in with radius $1$. So I think I have to modify a little this equation for something which radius varies and annulate in zero. If I put $\sin(\frac{1}{x})$ I got an infinity of points which annulate in zero, am I on the right way??
 A: One way to solve this problem is to construct a function $f:[0,\infty]\to\mathbb{R}$ which is $C^k$ and satisfies $$f(a_n)=a_n^p\tag{1}$$
where $a_k\to 0$ and $p>0$ is some real number. As you have suggested, we could try $\sin({1/r})$, but the problem with this function is the origin, i.e. this function satisfies $(1)$ for $p=1$ and because of this we have that the field $$
 \left\{ \begin{array}{ccc}
 x'=y+x\left(\sin{\left(\frac{1}{x^2+y^2}\right)}-(x^2+y^2)\right) &\mbox{ } \\
  y'=-x+y\left(\sin{\left(\frac{1}{x^2+y^2}\right)}-(x^2+y^2)\right) &\mbox{}
       \end{array} \right.
$$
has a sequence of limit cycles converging to $0$, but it is not $C^k$ at the origin. To remedy this, we note that the function $$f_k(r)=r^{2k+1}\sin{(1/r)},\ k=1,2,...$$ is $C^k$. Moreover, if we choose $p=2k+1$ then, for each $k$, there exist a sequence (depending on $k$) $a_n\to 0$ satisfying $(1)$. We conclude from this analysis that if $r=x^2+y^2$ then 
$$
 \left\{ \begin{array}{ccc}
 x'=y+x\left(r^{2k+1}\sin{\left(\frac{1}{r}\right)}-r^{2k+1}\right) &\mbox{ } \\
  y'=-x+y\left(r^{2k+1}\sin{\left(\frac{1}{r}\right)}-r^{2k+1}\right) &\mbox{}
       \end{array} \right.
$$
satisfies the desired requiremnts.
