What explains this bizarre behavior?

The answers of mercio and Mark Bennet explain how to find the antecedents of $1$, $0$, and $-1$. This answer may help to visualize the structure of the set.

The horizontal axis represents $x_0$. The black horizontal line segments at height $1$ indicate intervals on which the limiting value of $x_n$ is $+\infty,$ while the black horizontal line segments at height $-1$ indicate intervals on which the limiting value of $x_n$ is $-\infty.$ The colored dots at heights $\pm1$ indicate values of $x_0$ for which the fixed points $\pm1$ are reached, while the colored dots on the horizontal axis represent values of $x_0$ for which the fixed point $\infty$ is reached (by first reaching $0$). Redder colors indicate fixed points that are reached in a small number of iterations; bluer colors indicate fixed points that are reached in a larger number of iterations. The image is animated. If the animation has stopped, it may be necessary to reload the page.

Some data:

$x_0$ reaching fixed points $(1,\infty,-1)$ after $1$ step: $$\left( \begin{array}{ccc} -0.5 & 0. & 0.5 \\ \end{array} \right)$$

$x_0$ reaching fixed points $(1,\infty,-1)$ after $2$ steps: $$\left( \begin{array}{ccc} -0.84307 & -0.707107 & -0.59307 \\ 0.59307 & 0.707107 & 0.84307 \\ \end{array} \right)$$

$x_0$ reaching fixed points $(1,\infty,-1)$ after $3$ steps: $$\left( \begin{array}{ccc} -0.948618 & -0.905646 & -0.870752 \\ -0.574217 & -0.552092 & -0.527083 \\ 0.527083 & 0.552092 & 0.574217 \\ 0.870752 & 0.905646 & 0.948618 \\ \end{array} \right)$$

$x_0$ reaching fixed points $(1,\infty,-1)$ after $4$ steps: $$\left( \begin{array}{ccc} -0.982971 & -0.968882 & -0.957545 \\ -0.865086 & -0.858475 & -0.851051 \\ -0.587509 & -0.582428 & -0.577977 \\ -0.522169 & -0.516059 & -0.508662 \\ 0.508662 & 0.516059 & 0.522169 \\ 0.577977 & 0.582428 & 0.587509 \\ 0.851051 & 0.858475 & 0.865086 \\ 0.957545 & 0.968882 & 0.982971 \\ \end{array} \right)$$

$x_0$ reaching fixed points $(1,\infty,-1)$ after $5$ steps: $$\left( \begin{array}{ccc} -0.994334 & -0.989663 & -0.985915 \\ -0.955713 & -0.953578 & -0.951185 \\ -0.869077 & -0.86755 & -0.866214 \\ -0.849598 & -0.847795 & -0.845616 \\ -0.591285 & -0.589765 & -0.588514 \\ -0.577225 & -0.576336 & -0.575323 \\ -0.52566 & -0.524341 & -0.52317 \\ -0.507143 & -0.505222 & -0.502849 \\ 0.502849 & 0.505222 & 0.507143 \\ 0.52317 & 0.524341 & 0.52566 \\ 0.575323 & 0.576336 & 0.577225 \\ 0.588514 & 0.589765 & 0.591285 \\ 0.845616 & 0.847795 & 0.849598 \\ 0.866214 & 0.86755 & 0.869077 \\ 0.951185 & 0.953578 & 0.955713 \\ 0.985915 & 0.989663 & 0.994334 \\ \end{array} \right)$$

Explanation: Taking the first row of the last table, $$\begin{pmatrix} -0.994334 & -0.989663 & -0.985915\end{pmatrix},$$ as an example,

• $x_0=-0.994334$ implies $x_n=1$ for $n\ge5;$
• $x_0=-0.989663$ implies $x_n=\infty$ for $n\ge5$ since $x_4=0;$
• $x_0=-0.985915$ implies $x_n=-1$ for $n\ge5;$
• for $x_0\in(-0.994334,-0.989663),$ we have $\displaystyle\lim_{n\to\infty}x_n=+\infty;$
• for $x_0\in(-0.989663,-0.985915),$ we have $\displaystyle\lim_{n\to\infty}x_n=-\infty.$

The decimal numbers in this discussion should, of course, be replaced by the algebraic numbers they approximate.

The recurrence has fixed points at 1 and -1.

If $x_n \lt -1$ then the sequence becomes increasingly negative.

If $x_n \gt 1$ then the sequence becomes increasingly positive.

The sequence crashes if $x_n=0$

The behaviour outside the interval $(-1,1)$ is clear, so we don't need long strings of values to explain it.

The way to understand what is going on is to see that the interval $(-1,1)$ can be separated into subintervals in which the values exhibit the same behaviour. The behaviour changes when we pass through a critical point - and these are just the points at which the sequence eventually terminates at $0$ or $\pm1$.

So, for example, if $x_n \in (-0.5,0)$ we find that $x_{n+1} \gt 1$ and $x_n \in (0,0.5)$ gives $x_{n+1} \lt -1$. The behaviour changes as we pass through $0$.

We note that if $x_n=\pm0.5$ then $x_{n+1}=\mp1$, so these are critical points. Also if $x_n=\pm \frac {\sqrt 2} 2$ then $x_{n+1}=0$.

If $x=a$ is a critical point, we get the next one(s) - the point(s) which map onto $a$ - by solving $2x-\cfrac 1 x=a$, this is equivalent to $$2x^2-ax-1=0$$ with solutions $$\frac {a\pm\sqrt{a^2+8}}4$$

Extending the function $f : x \mapsto 2x- \frac 1x$ to the circle $\Bbb R \cup \{\infty\}$ by defining $f(0) = f(\infty) = \infty$, $f$ is a $2$-to-$1$ function with $3$ fixpoints, one attractive ($\infty$), and two repulsive ($1,-1$).

So, for most starting reals, the generated sequence will "converge" to $\infty$. Since $f$ is continuous, the set of reals that generate a sequence diverging to $+ \infty$ is an open subset, as well as the set of reals that generate a sequence diverging to $- \infty$.

those two open sets "come in contact" at the real numbers that generate a sequence eventually going to $0$ and then that stays on $\infty$. Here, you can plainly see that there is an $x \in (0.87 ; 0.88)$ such that $f^{6}(x) = 0$.

To find the precise value of that $x$, you need to compute the successive antecedents of $0$ by $f$ that lay in-between the two sequences.

$f^{-1}(y) = \frac{y \pm \sqrt {y^2+8}}4$, so the first one is $\frac {\sqrt 2}2$ (betwwen the two $X(5)$ values), then $\frac {\sqrt 2 + \sqrt {34}}8$ (between the two $X(4)$ values), and you can go on until you reach the incriminating $x \in (0.87;0.88)$.

Also, the set of all the antecedents of $0$ has accumulation points (starting with $1$ and $-1$, but they could be more), so it's very possible to find many incriminating values between the two initial values, by going back long enough in the antecedents tree.