Type of an isometry of a $\delta$-hyperbolic space Let $(X,d)$ be proper geodesic $\delta$-hyperbolic metric space.
Let $\gamma \in Isom(X)$. Denote by $\partial X$ the boundary at infinity of $X$ (which is invariant of base-point).
Let $x\in X$. We say that $\gamma$ is:
1) elliptic, if the orbit of x under $\gamma$, $o(x) := \{ \gamma ^n.x |n\in Z\}$ is bounded;
2) parabolic, if $o(x)$ has exactly one accumulation point in $\partial X$;
3) hyperbolic, if $o(x)$ has exactly two accumulation points in $\partial X$.
It turns out that this definition is independent of the choice of $x$.
My problem is to show that $\gamma$ is elliptic, parabolic, or hyperbolic.
For this, assume it is not elliptic nor parabolic. By Arzela-Ascoli we can conclude, in this case, that $o(x)$ has at least two accumulation points in $\partial X$. So If we show that $o^+(x) := \{ \gamma ^n.x |n\in N\}$ has exactly one accumulation point in $\partial X$, then the result follows (by observing that the orbit of $x$ under $\gamma^{-1}$ is $o(x)$).
To show this, we observe that again by Arzela-Ascoli $o^+(x)$ has at least one accumulation point in $\partial X$, but i can't see why it is unique.
Any body can help me in this? thanks.
 A: Here is a proof following the book by Ghys and de la Harpe:
In what follows, $X$ is a $\delta$-hyperbolic geodesic metric space. 
Lemma 1. Suppose that $C\in {\mathbb R}$, $x_n, y_n\in X$ are such that $d(x_n, y_n)\le C$ for all $n$. Then $(x_n)$ converges to a point $\xi$ in $\partial X$ if and only if  $(y_n)$ converges to the same point $\xi$. 
I will omit the proof, see if you can prove it yourself.
Lemma 2. Suppose that $g$ is an isometry $X$, $x\in X$ and the sequence $x_n=g^n(x)$ accumulates to a point $\xi\in \partial X$. Then $\xi$ is fixed by $g$. 
Proof. Suppose that $n_j$ is a strictly increasing sequence of natural numbers such that the subsequence $(x_{n_j})$ converges  to $\xi$. Lemma 1 implies that the sequence
$$
(x_{n_j+1})= (g^{n_j} g(x))
$$
also converges to $\xi$. But the multiplication by $g$ transforms the sequence $(x_{n_j})$ to $(x_{n_j+1})$. Hence, $g$ fixes the limit point $\xi$ of both sequences. qed 
Corollary. The accumulation sets $A_\pm$ of the sequences $(g^n(x)), (g^{-n}(x))$ is fixed by $g$ pointwise. In other words, $g$ fixes the limit set $A$ of the group $G=\langle g\rangle$ pointwise. 
Lemma 3. The limit set $A$ has cardinality $\le 2$. 
Proof. Suppose to the contrary that $A$ contains at least three distinct points $\xi_1, \xi_2, \xi_3$. These points are ideal vertices of an ideal triangle $T$ in $X$ (formed by three complete geodesics $c_1, c_2, c_3$ asymptotic to the points $\xi_1, \xi_2, \xi_3$). Since $g$ fixes $A$ pointwise, it "almost preserves" $T$, i.e. for each side $c_i$ of $T$, for every $n\in {\mathbb Z}$, the Hausdorff distance between $g^n(c_i)$ and $c_i$  is at most $2\delta$ (or some other function of $\delta$ depending on your definition of hyperbolicity). Let $c$ denote a center of $T$, i.e. a point within distance $\le 10\delta$ (or some other function of $\delta$) from all three sides of $T$. Then for every $n\in {\mathbb Z}$, $g^n(c)$ is also a center of $T$. Since the distance between any two centers of an ideal triangle is at most, say, $100\delta$, it follows that the $G$-orbit of $c$ is bounded. But then $g$ is elliptic and cannot have any accumulation points in $\partial G$. A contradiction. qed 
Lastly, you are really handicapping yourself by not reading French. If you are seriously planning to study geometric group theory, my suggestion is to make an effort learning to read mathematical French. Unlike, say, mathematical German, it is reasonably close to the mathematical English. 
