
Le fait que $u_{n+1}=\sqrt{2+u_n}$ nous dit que $u_{n+1}=f(u_n)$, avec $f(x)=\sqrt{2+x}$. Il faut donc tracer la courbe représentative de la fonction $f$ (en rouge ci-dessous) ainsi que la droite d'équation $y=x$ (en orange ci-dessous) pour transformer les images (en ordonnées) en abscisses.

\begin{tikzpicture}[baseline=3.8cm,xscale=1,yscale=1]
\draw[dotted,gray,opacity=0.5] (-1,-1) grid[step=0.2] (15,4);
\draw[gray,opacity=0.5] (-1,-1) grid (15,4);
\draw[thick,->,>=latex] (-1,0) -- (15,0);
\draw[thick,->,>=latex] (0,-1) -- (0,4);
\clip (-1,-1) rectangle (15,4);
\draw[thick,orange] plot[domain=-1:15] (\x,\x) node[below left,rotate=45.0,scale=0.7,outer xsep=3mm] {$y=x$};
\draw[thick,red] plot[domain=0:13,samples=100] (\x,{sqrt(\x+2)});
% Graduation des axes
\draw[thick] (-1,0.1/1) -- (-1,-0.1/1) node[below,outer sep=2mm] {-1};
\draw[thick] (1,0.1/1) -- (1,-0.1/1) node[below,outer sep=2mm] {1};
\draw[thick] (2,0.1/1) -- (2,-0.1/1) node[below,outer sep=2mm] {2};
\draw[thick] (3,0.1/1) -- (3,-0.1/1) node[below,outer sep=2mm] {3};
\draw[thick] (4,0.1/1) -- (4,-0.1/1) node[below,outer sep=2mm] {4};
\draw[thick] (5,0.1/1) -- (5,-0.1/1) node[below,outer sep=2mm] {5};
\draw[thick] (6,0.1/1) -- (6,-0.1/1) node[below,outer sep=2mm] {6};
\draw[thick] (7,0.1/1) -- (7,-0.1/1) node[below,outer sep=2mm] {7};
\draw[thick] (8,0.1/1) -- (8,-0.1/1) node[below,outer sep=2mm] {8};
\draw[thick] (9,0.1/1) -- (9,-0.1/1) node[below,outer sep=2mm] {9};
\draw[thick] (10,0.1/1) -- (10,-0.1/1) node[below,outer sep=2mm] {10};
\draw[thick] (11,0.1/1) -- (11,-0.1/1) node[below,outer sep=2mm] {11};
\draw[thick] (12,0.1/1) -- (12,-0.1/1) node[below,outer sep=2mm] {12};
\draw[thick] (13,0.1/1) -- (13,-0.1/1) node[below,outer sep=2mm] {13};
\draw[thick] (14,0.1/1) -- (14,-0.1/1) node[below,outer sep=2mm] {14};
\draw[thick] (15,0.1/1) -- (15,-0.1/1) node[below,outer sep=2mm] {15};
\draw[thick] (0.1/1,-1) -- (-0.1/1,-1) node[left] {-1};
\draw[thick] (0.1/1,1) -- (-0.1/1,1) node[left] {1};
\draw[thick] (0.1/1,2) -- (-0.1/1,2) node[left] {2};
\draw[thick] (0.1/1,3) -- (-0.1/1,3) node[left] {3};
\draw[thick] (0.1/1,4) -- (-0.1/1,4) node[left] {4};
\draw[dashed,purple] (10,0) node[below,scale=0.5,outer ysep=3mm] {$u_{0}$} -- (10,3.4641016151377544);
\draw[dashed,purple] (10,3.4641016151377544) -- (3.4641016151377544,3.4641016151377544);
\draw[dashed,purple] (3.4641016151377544,3.4641016151377544) -- (3.4641016151377544,0);
\draw[dashed,purple] (3.4641016151377544,0) node[below,scale=0.5,outer ysep=3mm] {$u_{1}$} -- (3.4641016151377544,2.3375417889607353);
\draw[dashed,purple] (3.4641016151377544,2.3375417889607353) -- (2.3375417889607353,2.3375417889607353);
\draw[dashed,purple] (2.3375417889607353,2.3375417889607353) -- (2.3375417889607353,0);
\draw[dashed,purple] (2.3375417889607353,0) node[below,scale=0.5,outer ysep=3mm] {$u_{2}$} -- (2.3375417889607353,2.0826765925031987);
\draw[dashed,purple] (2.3375417889607353,2.0826765925031987) -- (2.0826765925031987,2.0826765925031987);
\draw[dashed,purple] (2.0826765925031987,2.0826765925031987) -- (2.0826765925031987,0);
\node[below,scale=0.5,outer ysep=3mm,purple] at (2.0826765925031987,0) {$u_{3}$};
%
\end{tikzpicture}

\medskip

On peut alors conjecturer que la limite de cette suite est l'abscisse du point d'intersection de la courbe avec la droite, soit la solution de l'équation :
\begin{align*}
x=\sqrt{2+x} & \iff x^2 = 2+x\\
& \iff x^2-x-2=0\qquad,\qquad\Delta=9\\
& \iff x = 2.
\end{align*}

\begin{attention}
Ce n'est qu'une conjecture... On pourra le démontrer à l'aide du théorème du point fixe (voir chapitre sur la continuité).
\end{attention}
