Let $x_1 = 2$. For $j \geq 1$, set $$x_{j+1} = x_j - \frac{x_j^2-2}{2x_j}$$
Show that the sequence $\{x_j\}$ is decreasing and bounded below. What is its limit?
Idea: We can show that the sequence is decreasing by an induction argument. We assume that $\sqrt{2} < x_i \leq 2$ for all $i \leq k$ and show that $\sqrt{2} < x_{k+1} \leq 2$. This can be easily shown since $x_{j+1} - x_j = - \frac{x_j^2-2}{2x_j}$, and the right hand side of the equation is less than $0$ and greater than or equal to $-\frac{1}{2}$.
By similar idea, it is not hard to show that the sequence converges by the Cauchy criterion. But how do we find the limit of the sequence and show that the sequence converges to this limit?
If a sequence defined by $x_{n+1} = f(x_n)$ converge, it converge to a fixed point of $f$.
Hence the limit verify
$L = L - \frac{L^2-2}{2L}$
And it follow that $L=\sqrt{2}$
Your sequence is basically the Newton's method applied to the function $f(x)=x^2-2$