= Solution
Write $v=\dot x$. The <simple pendulum> becomes $\dot x=v$, $\dot v=-\sin x$. Its <equilibria> are $(k\pi,0)$. The <linearization> is $\begin{pmatrix}0&1\\-\cos(k\pi)&0\end{pmatrix}$. For even $k$ its <eigenvalues> are $\pm i$, and for odd $k$ they are $\pm1$. The conserved energy
$$
H=\frac{v^2}{2}+1-\cos x
$$
has a strict local minimum at each even multiple of $\pi$, proving nonlinear <Lyapunov stability> there. The odd multiples are saddles and are unstable. Therefore
$$
\boxed{(2k\pi,0)\text{ are stable centres};\qquad((2k+1)\pi,0)\text{ are unstable saddles}.}
$$
The stable centres are not asymptotically stable because energy is conserved.
The saddle energy is $H=2$, so the <separatrix> is determined by
$$
\boxed{\frac{\dot x^2}{2}+1-\cos x=2,\qquad
\dot x=\pm2|\cos(x/2)|.}
$$
Energy levels $0<H<2$ form closed libration curves around the centres; levels $H>2$ have velocity of constant sign and describe rotations. The separatrices join neighboring saddles in the unwrapped $(x,v)$ plane. On the angle cylinder, where $x$ is identified modulo $2\pi$, they are homoclinic to the single saddle.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ia/paper-4-pendulum-phase-portrait.png]
{title=Pendulum energy levels, stable centres, unstable saddles and the separatrix}
At $x=0$ the positive separatrix has $\dot x=2$. Before reaching the neighboring saddle at $x=\pi$, the velocity is $2\cos(x/2)>0$, so the time to angle $a<\pi$ is
$$
t(a)=\int_0^a\frac{dx}{2\cos(x/2)}
=\log\left(\sec(a/2)+\tan(a/2)\right).
$$
As $a\uparrow\pi$, $\cos(x/2)\sim(\pi-x)/2$. The integrand is therefore asymptotic to $1/(\pi-x)$, whose improper integral diverges logarithmically. Thus the <pendulum separatrix approach takes infinite time>:
$$
\boxed{T=\int_0^\pi\frac{dx}{2\cos(x/2)}=\infty.}
$$
Starting with velocity $-2$ gives the reflected approach to $x=-\pi$ and the same infinite time. The velocity tends to zero but never reaches zero at a finite time.
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