## Problem 3.

Let the sequence of real numbers $\left(x_{n}\right)_{n \geq 0}$ be defined by

$$
\begin{aligned}
& x_{0}=0 \\
& x_{n}=1+\sin \left(x_{n-1}-1\right),(\forall) n \geq 1
\end{aligned}
$$

Study the convergence of the sequence and then determine its limit.
## Solution:

Since $x_{n}-1=\sin \left(x_{n-1}-1\right),(\forall) n \geq 1$

If we denote $a_{n}=x_{n}-1,(\forall) n \geq 1$, then $a_{n}=\sin a_{n-1},(\forall) n \geq 1$.

Since $a_{0}=-1$, we deduce that $a_{n}<0,(\forall) n \geq 1$.

$\left(a_{n}\right)_{n \geq 0}$ is increasing

Thus convergent and its limit exists $\lim _{n \rightarrow \infty} a_{n}=a, a \in R$

It follows that $a=\sin a$, (1p).

Therefore $a=0$, and $\lim _{n \rightarrow \infty} x_{n}=1$ .(1p).