Given: $f \in C^2[a,b]$ with $f(r) = 0$, $f'(r) \neq 0$ for $r \in (a,b)$.

To Prove: There exists $\delta > 0$ such that if $|x_0 - r| < \delta$, then Newton's iteration converges quadratically with: $\lim_{n \to \infty} \frac{|x_{n+1} - r|}{|x_n - r|^2} = \frac{|f''(r)|}{2|f'(r)|}$

Proof:

Define the error $e_n = x_n - r$. Newton's iteration is: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

Therefore: $e_{n+1} = x_{n+1} - r = x_n - \frac{f(x_n)}{f'(x_n)} - r = e_n - \frac{f(x_n)}{f'(x_n)}$

Since $f(r) = 0$, Taylor's theorem with remainder gives: $f(x_n) = f(r) + f'(r)(x_n - r) + \frac{f''(\xi_n)}{2}(x_n - r)^2$ $= f'(r)e_n + \frac{f''(\xi_n)}{2}e_n^2$

where $\xi_n$ is between $x_n$ and $r$.

Substituting into the error equation: $e_{n+1} = e_n - \frac{f'(r)e_n + \frac{f''(\xi_n)}{2}e_n^2}{f'(x_n)}$

$= e_n - \frac{f'(r)}{f'(x_n)}e_n - \frac{f''(\xi_n)}{2f'(x_n)}e_n^2$

$= e_n\left(1 - \frac{f'(r)}{f'(x_n)}\right) - \frac{f''(\xi_n)}{2f'(x_n)}e_n^2$

Now, since $f'$ is continuous and $f'(r) \neq 0$, there exists $\delta_1 > 0$ such that $|x_n - r| < \delta_1$ implies $f'(x_n)$ has the same sign as $f'(r)$ and $|f'(x_n)| \geq |f'(r)|/2$.

For $|x_n - r| < \delta_1$, as $x_n \to r$, we have $f'(x_n) \to f'(r)$, so: $1 - \frac{f'(r)}{f'(x_n)} \to 0$

More precisely, by continuity: $\frac{f'(r) - f'(x_n)}{f'(x_n)} = O(|x_n - r|)$

The dominant term is: $e_{n+1} = -\frac{f''(\xi_n)}{2f'(x_n)}e_n^2 + O(e_n^3)$

Since $f''$ and $f'$ are continuous, $f''(\xi_n) \to f''(r)$ and $f'(x_n) \to f'(r)$ as $n \to \infty$. Therefore: $\frac{|e_{n+1}|}{|e_n|^2} \to \frac{|f''(r)|}{2|f'(r)|}$

This establishes quadratic convergence. The error bound follows: $|e_{n+1}| \leq \frac{M}{2m}|e_n|^2$ where $M = \sup|f''(x)|$ and $m = \inf|f'(x)|$ on a neighborhood of $r$.