The exercise goes as follows:
Exercise 5.25:
25. Suppose $f$ is twice differentiable on $[a,b]$, $f(a)\lt 0$, $f(b)\gt 0$, $f'(x)\geq \delta\gt 0$, and $0\leq f''(x)\leq M$ for all $x\in[a,b]$. Let $\xi$ be the unique point in $(a,b)$ at which $f(\xi) = 0$.
Complete the details in the following outline of Newton's method for computing $\xi$.
(a) Choose $x_i\in (\xi,b)$, and define $\{x_n\}$ by $$x_{n+1}=x_n - \frac{f(x_n)}{f'(x_n)}.$$ Interpret this geometrically, in terms of a tangent to the graph of $f$.
(b) Prove that $x_{n+1}\lt x_n$ and that $$\lim_{x\to\infty} x_n = \xi.$$
I have problems regarding part b. I have managed to prove that $x_{n+1}<x_{n}$ and that $\xi<x_{n}<b$ (for all n). I also know that there must be a limit since it is a monotone sequence and it is bounded. However, I do not know how to prove that the desired limit is $\xi$.
I have seen some argument of the sort that if the limit is $\eta$, then we should have that $\eta = \eta - f(\eta)/f'(\eta)$, which would imply $f(\eta)=0$ and so $\eta = \xi$. But I do not understand why is this valid since the limit does not necessarily imply that the sequence $x_{n}$ contains $\eta$.
How could I prove the remaining part of the exercise rigorously?
Consider that $\lim\limits_{n \to \infty} x_n = \lim\limits_{n \to \infty} x_{n + 1} = \lim\limits_{n \to \infty} x_n - \frac{f(x_n)}{f'(x_n)} = \lim\limits_{n \to \infty} x_n - \lim\limits_{n \to \infty} \frac{f(x_n)}{f'(x_n)}$.
Therefore, $\lim\limits_{n \to \infty} \frac{f(x_n)}{f'(x_n)} = 0$. Since $f'(x_n) \geq \delta > 0$ for all $n$, we see that $\lim\limits_{n \to \infty} f(x_n) = 0$. Since $f$ is continuous, this gives us $f(\lim\limits_{n \to \infty} x_n) = 0$.
Note that you actually can weaken the hypotheses further. You only need $f$ to be convex and for there to be some $\delta > 0$, $K$ such that $\delta \leq f'(x) \leq K$ for all $x$. Convexity is enough to show that $f(x_n) \geq 0$ for all $x$, while the boundedness of $f'$ is enough to go from $\lim\limits_{n \to \infty} \frac{f(x_n)}{f'(x_n)} = 0$ to $\lim\limits_{n \to \infty} f(x_n) = 0$.