If a vector field $V$ has a semi stable limit cycle then there is no any volum form $\Omega$ such that $Div_{\Omega} X=C$, a constant.
Explanation:
A semi stable limit cycle is an isolated periodic orbit which is attractor from the interior and repellor from the exterior(Or attractor from exterior and repellor from interior).
Because if divergence is a non zero constant, say positive(negative, resp.) so every closed orbit $\gamma$ is strongly hyperbolic hence is a repellor (attractor, resp.) limit cycle so it attracts a FULL neighborhood of $\gamma$ in negative or positive time. Any such closed orbit can not be a semi stable limit cycle. Recall that a closed orbit $\gamma$ is called a strongly atractive hyperbolic if the divergence of the field is negative at all point of $\gamma$ and is strongly repellor hyperbolic if the divergence is postive at all points of $\gamma$.
On the other hand if $C=0$ then the flow is volume preserving so this contradicts to the fact that a half neighborhood of a closed orbit $\gamma$ tends to $\gamma$ as times goes to $\pm \infty$
Note: Semi stable limit cycles exists via concrte examples:
For example the following system has a semi stable limit cycle $\gamma: x^2+y^2=1$:
$$\begin{cases} x'=y+x(x^2+y^2-1)^2\\y'=-x+y(x^2+y^2-1)^2 \end{cases}$$
but they are very sensitive in the sense that they disappear or they would be twice by very small perturbation. In the foliation language, a semi stable limit cycle of a vector field on a surface is a closed leaf whose holonomy map $h$ satisfies either $h(x)-x>0 \quad\forall x\in (-\epsilon, \epsilon)\setminus \{0\}$ or $h(x)-x<0 \quad \forall x\in (-\epsilon, \epsilon)\setminus \{0\}$ for $\epsilon$ sufficiently small.
For a related post on semi stable limit cycle see this question on quadratic systems