Considering IVP:
$\dot{x} = \cos(x) +1, \; x(0) = 0.$
I should compute the omega and alpha limit of the given initial point.
So I first found equilibria, which in this case is: $ x = k \pi, \; k \in \mathbb{Z}$, and this equilibrium is not hyperbolic.
But I am not sure how to continue further?
Equilibrium points would actually be $x=k\pi$, with $k$ odd (if $k$ is even, $\cos(x)=1$).
For this reason, the lines $y=-\pi$, $y=\pi$ are trajectories of the system.
It is well known trajectories cannot intersect each other, so the solution starting at $(0,0)$ will always be contained in the band bounded by those two lines.
Now, if $x(t)$ is that solution, $x'(t)=\cos(x(t))+1>0$, because $-\pi<x(t)<\pi$.
Thus, the solution is increasing and bounded above, so it has a limit at $t\to\infty$. But this limit can only be an equilibrium point, and so $x(t) \uparrow \pi$, and $\omega(0)=\{\pi\}$.
Analogously, as the solution is increasing and bounded below, it has a limit at $t\to-\infty$, and this limit can only be $-\pi$, so $\alpha(0)=\{-\pi\}$.
