both $c$ and $d$ to $\infty$ (i.e. the limit of $M_n\circ i_n$). Indeed, if we take another normalization, i.e. another sequence of Moebius transformation $N_n$ that maps now $a$ to $0$, $c$ to $\infty$ and $d$ to $1$, one can see that this process would give a different limit for $[i_n]$.

It is an easy exercise to check that if we choose another sequence of representative such that the image of three elements of $X$ are constant we have a limit with differs only from the two previous one by a Moebius translation. So a first idea is to think that the limit that we define for $[i_n]$ is going to be the collection of these two possible limits.

The second idea is to remark that these two limits are related. Indeed, if for example $M_n\circ i_n(d)=n$ then we can compute that $N_n\circ M_n^{-1}(z)= z/n$ and it follows that $N_n\circ i_n(b)=1/n\to0=N_n\circ i_n(a)$. Hence this collection of limits comes with some rigidity. This is why we introduce trees: in order to give a simple visualization of this rigidity. In this example we identify the limit of $[i_n]$ to a tree of sphere (that we will define later) as on Figure \[dmpp\].

