Simple limit of a sequence defined recursively For each $ \ N \in \mathbb{R} \ $ and each $ \ t \in \mathbb{R} \, $, let $ \ S(N,t) = (X_n)_{n \in \mathbb{N}^*} \in \mathbb{R}^{\mathbb{N}^*} \ $ be a sequence such that $ \ X_1 = t \ $ and, $\forall n \in \mathbb{N}^*$, $$X_{n+1} = X_n + \frac{N - X_n^2}{2 X_n} \ \ . $$ I want to know for which values of $N$ and $t$ the sequence converges and the value of the limits $$L(N,t) = \lim \, S(N,t) = \lim_{n \to \infty} X_n \ \ . $$
I have a feeling that, for $ \ 0<t<N< \infty \ $ the sequence converges to $ \sqrt{N}$, because I have an intuitive idea of some areas of a square I have drawn (picture below). The square has area $ \ N = (a+t)^2 \, $, with sides of lenght $ \ a+t = \sqrt{N}$, where $ \ 0<a<t \, $. So we have $ \ N = t^2 + 2at + a^2 \, $. In my first approximation I simply forget the small $ \, a^2 \, $ and get $ \ N = t^2 + 2at \ \Rightarrow \ a = \frac{N-t^2}{2t} \, $. Thus, my first approximation was a square of sides $ \ X_1 = t \ $ and my second approximation was a square of sides $ \ X_2 = X_1 + a \ $ and I proceed in this fashion to get the sequence above. But this is just a feeling and I want to know if this process converges and if I can generalize this to any $ \ N \in \mathbb{R} \ $ and any $ \ t \in \mathbb{R} \, $.
Thanks.

 A: Indeed, the sequence converges to $\sqrt N$.
The sequence you're looking at is actually reasonably well-known: it is the sequence of Newton's method iterations for finding a root of the function $f(x) = x^2 - N$. Your recurrence can be rewritten as $X_{n+1} = X_n - \frac{f(X_n)}{f'(X_n)}$, which is just a single step of Newton's method.
If you don't recognize this sequence, then it's easy to see that if it converges to anything, it must converge to $\pm \sqrt N$. Write the recurrence as $X_{n+1} = g(X_n)$. Then if $X_n \to X$ as $n \to \infty$, we must have $g(X_n) \to g(X)$ as $n \to \infty$. But the sequence $g(X_n)$ is just $X_n$, one iteration ahead. So $X$ must satisfy $g(X) = X$. In this case, $$X = X - \frac{N - X^2}{2X} \quad\Rightarrow\quad \frac{N - X^2}{2X} = 0 \quad\Rightarrow\quad X^2 = N.$$
Actually showing convergence is tricky. One way is to observe that if $X_n > \sqrt N$, then $X_{n+1} > \sqrt N$ but $X_{n+1} < X_n$ (which just takes some more algebra to show), and if $0 < X_n < \sqrt N$, then $X_{n+1} > \sqrt N$. So the sequence $t = X_1, X_2, X_3 \dots$ will keep getting smaller and smaller while staying above $\sqrt N$. It must converge to something and that something can only be $\sqrt N$.
