For given smooth vector field $X$ on Riemannian manifold,solution of $\dot c=X(c)$ is the flow line or integral curve.
Then the point $\lim\limits_{t\rightarrow\infty} c_p(t)$ need not be contained in the flow line.But I think if the manifold is compact it will be contained in the flow line.Is it right ?
The vector field $X(\theta) = \sin \theta$ on the real line has zeros at integer multiples of $\pi$, and no flow line reaches a zero in finite time. Because $X$ is invariant under translation by $2\pi$, $X$ descends to the (compact) quotient, a circle. The character of the flow on the quotient is locally identical to the flow "upstairs". Particularly, no flow line reaches its limit in finite time.