Asymptotic frequency of nonlinear oscillator $\ddot x = -x-{\dot x}^3$ (speed cubed) A particle oscillates according to the equation
$\ddot x = -x-{\dot x}^3.$ The positive positions of the particle when it changes direction, $\dot x = 0$, are $x_1,x_2,\ldots$.

I want to show that
$$\frac{|x_n|-|x_{n+1}|}{|x_n|^3}$$ tends to a constant as $n\to \infty$.

Note that the energy $E = x^2 + \dot x^2 $ decreases with time, since $\dot E = -2\dot x ^4 < 0$. Thus eventually the $\dot x ^3$ perturbation term is very small and we indeed oscillate "almost perfectly" and expect $x_n,x_{n+1}$ to be very close. How can I show this quantitative relation?
 A: For what it's worth, here is a rough estimate for $n\gg 1$:
$$\begin{align} \frac{|x_n|-|x_{n+1}|}{|x_n|^3}
~\approx~&\frac{x_n^2-x_{n+1}^2}{2x_n^4}\cr 
~=~&\frac{E_n-E_{n+1}}{2E^2_n}\cr
~\approx~&-\frac{\langle \dot{E}\rangle}{E^2_{n+1/2}} \frac{T}{4}\cr
~\sim~&-\frac{\dot{E}_{n+1/2}}{E^2_{n+1/2}} \frac{T}{8}\cr
~=~&\frac{2\dot{x}^4_{n+1/2}}{\dot{x}^4_{n+1/2}}\frac{T}{4}\cr
~=~&\frac{T}{4}.
\end{align}$$
Notation: Note that the integer $n$ in this answer counts turning points, i.e points with zero velocity (whereas OP only counts positive turning points), i.e. in this answer $n$ increases with 2 per period $T\approx 2\pi$ (whereas OP increases $n$ by 1 per period.) Also note that the half-integer labelling denotes points with zero position.
A: Not sure if this will help, but motivated by your note that $\dot x^3$ is small, lets introduce a second time scale $\tau = \epsilon t$ and rescale the variable $x = \epsilon^\alpha y$. Plugging this in
$$ y_{tt} + 2 \epsilon y_{t\tau} + \epsilon^2 y_{\tau \tau} + y = -\epsilon^{2\alpha} (y_t + \epsilon y_\tau)^3. $$
We seek an expansion in $y = y^0 + \epsilon y^1 + \dots$. Lets choose $\alpha = 1/2$ so it enters in at first order of epsilon. Lets take $y(0) = 0$ and $y'(0) = 1$ The first two orders
\begin{align}
& \mathcal O(1): \quad y^0_{tt} + y^0 = 0 \\
& \mathcal O(\epsilon): \quad y^1_{tt} + y^1 = -2y^0_{t\tau} - (y_t^0)^3.
\end{align}
The first equation implies that $y^0 = A(\tau) \sin t$. To avoid secular terms at next order, we have the orthogonality condition $\langle y^0,-2y^0_{t\tau} - (y_t^0)^3 \rangle = 0$. This leads to our amplitude equation:
$$ A_\tau = -\frac{3}{8} A^3 $$
Solving this:
$$ y^0 = \frac{2 \sin t}{\sqrt{3\tau +4}}. $$
Which shows the $1/\sqrt t$ decay of the amplitude over (slow) time.
The points where $y_t(0) = \cos t = 0$ implies $t_n = \frac{\pi}{2} + \pi n$. Thus the limit
$$ \frac{|x_{n+1}| - |x_n|}{|x_n|^3} \to - \frac{3\pi}{8} \quad \text{as } n \to \infty. $$
A: Try to solve your second-order differential equation $\ddot x = -x-{\dot x}^3$ numerically and plot phase diagram $v(x)$ also $v(t)$. I love open-source Maxima Computer algebra System, cause it's free and does bunch of good work. So as for maxima - one can use function plotdf to analyse differential equation solution numerically. Full maxima code :
plotdf ( [v, -x-(v)^3], [x, v], [trajectory_at, 1, 0],[x, -1.5, 1.5], [v, -1.5, 1.5],[versus_t, 1],[tstep, 1], [nsteps, 1000])$

produces phase diagram :

and oscillations plot :

From both graphs can be seen than when $t \to \infty$ then $\delta x_{max} \to 0.$
In addition to that I want to say that many complex, non-linear and n-th order differential equations may not have analytical solutions, so numerical solving method can be the only one left in your hands. You can also try Runge-Kutta numerical ODE solving method.
EDIT
I was not able to find an analytical solution $x(t)$ to your given differential equation, cause maxima ODE packages has failed doing so.
However I have a gut-feeling that this solution takes form :
$$ x(t) = \frac {A \cdot \sin (\omega~t)}{\sqrt {~\omega~t~~}} $$
or something similar to it. Because if you draw such function you'll get exactly your differential equation behavior :

A: Perform an anti-rotation on $(x,\dot x)$, that is, consider new coordinates
$$
\begin{cases}
u(t)=x(t)\cos(t)-\dot x(t)\sin(t)
\\
v(t)=x(t)\sin(t)+\dot x(t)\cos(t)
\end{cases}
\implies
\begin{cases}
 \dot u(t)=-\sin(t)(\ddot x(t)+x(t))=\sin(t)\dot x(t)^3
\\
\dot v(t)=\cos(t)(\ddot x(t)+x(t))=-\cos(t)\dot x(t)^3
\end{cases}
$$
With $\dot x(t)=v(t)\cos(t)-u(t)\sin(t)$ the third power is
$$
\dot x(t)^3 = \cos^3(t)v^3-3\cos^2(t)\sin(t)uv^2+3\cos(t)\sin^2(t)u^2v-\sin^3(t)u^3
$$
If $(x,\dot x)$ is small, this gives a very slowly moving curve $(u,v)$. Replace $u,v$ with their averages over a period of $2\pi$ and integrate the equation-of-motion to find the average slope as
$$
\dot{\bar u}=-\frac38\bar u\bar v^2-\frac38\bar u^3
\\
\dot{\bar v} =-\frac38\bar v^3-\frac38\bar u^2\bar v 
$$
This means that these averaged quantities fall straight into the center, with $$\bar u(t)^2+\bar v(t)^2=\frac{r_0^2}{1+\frac34r_0^2t}$$ and thus
$$
\bar u(t)=\frac{u_0}{\sqrt{1+\frac34r_0^2t}}
\\
\bar v(t)=\frac{v_0}{\sqrt{1+\frac34r_0^2t}}
$$
Let's test the theory on a numerical example, using $x(0)=0.3$ and $\dot x(0)=0.1$ I get

As far as can be seen, the average behavior is well-caught.
For the specific limit asked about in the question, setting $(x(0),\dot x(0))=(x_n,0)$ in the averaging process, gives $x_{n+1}$ at $t=2\pi$. From the averaging formulas with $u_0=x_n$, $v_0=0$, $r_0=x_n$, one gets
$$
x_{n+1}-x_n=\frac{x_n}{\sqrt{1+\frac32\pi x_n^2}}-x_n=-\frac{\frac32\pi x_n^3}{\sqrt{1+\frac32\pi x_n^2}+1+\frac32\pi x_n^2}.
$$
So indeed the proposed limit exists as claimed, and as $x_n\to 0$ one gets
$$
\frac{x_{n+1}-x_n}{x_n^3}\to-\frac34\pi
$$
t = np.linspace(0,60,1000)
x = odeint(lambda x,t: [x[1],-x[0]-x[1]**3],[0.3,0.1],t,atol=1e-8,rtol=1e-10 )
x = x.T
ct, st = np.cos(t), np.sin(t)
u = x[0]*ct-x[1]*st
v = x[0]*st+x[1]*ct

ax[0].plot(t,u,t,v);
r02 = u[0]**2+v[0]**2
ax[1].plot(t,u**2+v**2,t,r02/(1+0.75*r02*t)); 
ax3.plot(u,v);

