\textbf{Part A}

We consider the function $f$ defined on the interval $[0; +\infty[$ by
$$f(x) = \sqrt{x+1}.$$
We admit that this function is differentiable on this same interval.

\begin{enumerate}
  \item Prove that the function $f$ is increasing on the interval $[0; +\infty[$.
  \item Prove that for every real number $x$ belonging to the interval $[0; +\infty[$:
  $$f(x) - x = \frac{-x^2 + x + 1}{\sqrt{x+1} + x}.$$
  \item Deduce from this that on the interval $[0; +\infty[$ the equation $f(x) = x$ admits as unique solution:
  $$\ell = \frac{1+\sqrt{5}}{2}.$$
\end{enumerate}

\textbf{Part B}

We consider the sequence $(u_n)$ defined by $u_0 = 5$ and for every natural number $n$, by $u_{n+1} = f(u_n)$ where $f$ is the function studied in part A. We admit that the sequence with general term $u_n$ is well defined for every natural number $n$.

\begin{enumerate}
  \item Prove by induction that for every natural number $n$, we have
  $$1 \leqslant u_{n+1} \leqslant u_n.$$
  \item Deduce from this that the sequence $(u_n)$ converges.
  \item Prove that the sequence $(u_n)$ converges to $\ell = \frac{1+\sqrt{5}}{2}$.
  \item We consider the Python script below:
\begin{verbatim}
from math import *
def seuil(n):
    u=5
    i=0
    while abs(u-l)>=10**(-n):
        u=sqrt(u+1)
        i=i+1
    return(i)
\end{verbatim}
We recall that the command $\mathbf{abs}(\mathbf{x})$ returns the absolute value of $x$.
  \begin{enumerate}
    \item[a.] Give the value returned by \texttt{seuil(2)}.
    \item[b.] The value returned by \texttt{seuil(4)} is 9. Interpret this value in the context of the exercise.
  \end{enumerate}
\end{enumerate}