Newton's Method expressed as a limit Yesterday in my AP Calc class, we discussed Newton's method ( $x = x_n - (f(x_n)/f'(x_n))$ ) of finding zeros of a function for the first time. He explained it as an algorithm, where each repeated iteration moves you closer to the actual value. The fact that the "solution" is what the expression approaches, I was wondering if it can be expressed as a limit. I was unsure as to whether the repeated iteration would make this impossible, but I thought it might be a neat idea. Any answers or ideas/thoughts about this are welcome :)
 A: Given a continuous function $f:\>[a,b]\to{\mathbb R}$ with $f(a)<0$ and $f(b)>0$ the intermediate value theorem tells you that the set $S:=\{x\>|\>a\leq x\leq b, \ f(x)=0\}$ is not empty. If $f$ is differentiable on $\>]a,b[\>$, and you can prove that $f'(x)>0$ for all $x$ in this interval then $f$ is strictly increasing on $[a,b]$, and the set $S$ contains exactly one element $\xi$, called the solution of $f(x)=0$ in $[a,b]$. If $f$ is even twice differentiable in $\>]a,b[\>$, and you can prove that $f''(x)>0$ in this interval, then there is a proof that Newton's method, started with $x_0:=b$ (this is important!), will converge to the unique point $\xi\in\>]a,b[\>$ with $f(\xi)=0$, i.e., that $\lim_{n\to\infty} x_n=\xi$.
But this is not how people work in practice. Given an equation $f(x)=0$ (even in a multivariate environment) they try to obtain a global (but "fuzzy")  overview of the set $S:=f^{-1}(0)$. Such an overview may indicate that there is a zero of $f$ "near" some point $x_0$ in the domain of $f$. They will then start a Newton process at $x_0$. If $x_0$ was "sufficiently near" to an actual zero $\xi$ of $f$ the process will converge fast to this point $\xi$, and this can be observed "at runtime".
