Prove that succession generated from an iterative function, converges to the root of the function. The problem of finding the root of the function $f(x)$, i.e. the value of $x^{\ast}$ such that $f(x^{\ast}) = 0$, can be reformulated as $x = g(x)$, and therefore the root $x^{\ast}$ will be the value $x^{\ast} = g(x^{\ast})$. Where:

*

*$g(x)$ is the Iterative Function;

*$x^{\ast}$ is fixed point of $g(x)$
With this reformulation, it is possible to approximate the root $x^{\ast}$, through an iterative procedure as:
$$x_{k+1} = g(x_k), \qquad k=0,1, \ldots$$
with $x_0$ assigned with an initial value.
Theorem:

If $g(x) : [a,b] → \mathbb R$ continuous and such that $a \le g(x) \le b \quad \forall x \in [a,b] \Rightarrow \exists $ a fixed point $x^{\ast} \in [a,b]$.
Also, if $g(x)$ is derivable in $[a,b]$, and $\exists \, 0 < \rho <1$ such that $|g'(x)| \le \rho \forall x  \in (a,b)$ then the fixed point is unique and the succession generated from $x_{k+1} = g(x_k)$  converges to $x^{\ast}$.

Proof:
We prove only the last part, about the convergence. Since $x^{\ast}$ is a fixed point of $g(x)$, and therefore $x^{\ast} = g(x^{\ast})$, we have:
$$\begin{array}{l}
|x_{k+1} - x^{\ast}| 
& \overset{1}{=} & |g(x_k) - g(x^{\ast})| \\
& \overset{2}{=} & |g \big (x^{\ast} + \left (x_{k} - x^{\ast} \right ) \big ) - g(x^{\ast})| \\
& \overset{3}{=} & |g'(c_k)|\cdot |x_k - x^{\ast}|
\end{array}
$$
we applied Taylor series, and $c_k \in I(x_k, x^{\ast})$, where $I(x_k, x^{\ast})$ is the interval containing $x_k$ and $x^{\ast}$.
Since $|g'(c_k)| \le \rho$ then we have:
$$|x_{k+1} - x^{\ast}| \le \rho|x_k - x^{\ast}|$$
and applying many times this inequality we obtain:
$$|x_k - x^{\ast}| \le \rho^k |x_0 - x^{\ast}| $$
and therefore:
$$\underset{k → \infty}{\lim} |x_k - x^{\ast}| \le \underset{k → \infty}{\lim} \rho^k |x_0 - x^{\ast}| = 0$$
and the convergence is verified.  $\square$

The above is from my textbook, it is the suggested proof, but I don't understand (for now), what is the reasoning applied in the passages from $\overset{2}{=}$ to $\overset{3}{=}$.
Please, can you help me to understand? Many thanks!
edit: (Probably, it is weird a proof, so if you think a different proof is better than this, you can post it but with detailed passages. Thanks.)
 A: Ok, I'll explain the proof. We are going to use the mean value theorem. It says that if a function $f$ is differentiable in an interval $(a,b)$ and continuous in $[a,b]$ then there exists a point $c\in (a,b)$ such that $f(b)-f(a)=f'(c)(b-a)$.
Now, in the theorem we want to prove we know that $x_{k+1}=g(x_k)$ for all $k\in\mathbb{N}$ and that $x^*$ is a fixed point of $g$. Also, $g$ is differentiable in $[a,b]$ and $|g'(x)|\leq\rho$ for all $x\in [a,b]$. So using the mean value theorem we know that for each $k\in\mathbb{N}$ there exists a point $c_k$ between $x_k$ and $x^*$ such that $g(x_k)-g(x^*)=g'(c_k)(x_k-x^*)$. Hence, for all $k\in\mathbb{N}$ we have:
$|x_{k+1}-x^*|=|g(x_k)-g(x^*)|=|g'(c_k)||x_k-x^*|\leq\rho |x_k-x^*|$
Note that this is true for every $k$. Hence we conclude that:
$|x_k-x^*|\leq\rho|x_{k-1}-x^*|\leq\rho^2|x_{k-2}-x^*|\leq...\leq\rho^k|x_0-x^*|$ for each $k\in\mathbb{N}$
Since $0<\rho<1$ we know that $\rho^k|x_0-x^*|\to 0$ when $k\to\infty$. And since $0\leq |x_k-x^*|\leq\rho^k|x_0-x^*|$ the squeeze theorem tells us that $x_k\to x^*$.
