= Solution
{c}
The <fixed-point iteration> stays inside $[0,7]$ because $G$ maps that interval to itself. A <fixed point> satisfies $\xi=\sqrt{2+\xi}$, so $\xi\ge0$ and
$$
\xi^2-\xi-2=(\xi-2)(\xi+1)=0.
$$
Only $\xi=2$ is admissible; $-1$ fails the unsquared equation. With $q=1/(2\sqrt2)$, the <mean value theorem> gives the <error bound>
$$
|x_{k+1}-2|=|G(x_k)-G(2)|\le q|x_k-2|.
$$
Induction yields $|x_k-2|\le q^k|x_0-2|\le5q^k\to0$, since $0<q<1$. Therefore, for every allowed starting value,
$$
\boxed{\lim_{k\to\infty}x_k=2.}
$$
This estimate proves convergence directly, without assuming in advance that the sequence has a limit.
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