Sequence of rationals converging to $\sqrt{2}$ via Newton's method That $a_{n+1}=\frac{a_n}{2} + \frac{1}{a_n}$, with $a_1=1$, is a sequence of rationals converging to $\sqrt{2}$ is well known.
I recently saw that the construction of this sequence can be done using Newton's method for finding solutions of $f(x)=0$ for $f(x)=x^2-2$. But I did not understand the steps rigorously.
(1) Let $x_0$ be first approximate root of $f(x)=x^2-2$. 
(2) Linear approximation at $x_0$: the equation of (tangent) line (at $x_0$) is $y=f(x_0) + (x-x_0)f'(x_0)$. 
(3) This line intersects $x$-axis and take the $x$-intersept to be next approximate root, so we solve for $y=0$. We get $x=x_0-\frac{f(x_0)}{f'(x_0)}$; declare this to be next approximation, so call this $x_1$. 
(4) Repeat the process (3) with $x_0$ replaced by $x_1$, and declare $x_2=x_1 -\frac{f(x_1)}{f'(x_1)}$. 
Now, if we take $x_0=2$, and do above iteration, we get the sequence of rationals stated in the beginning.
Q.1 Can we take other approximate root to be any other integer or rational number? What is the importance of $2$ taking as first approximate root? 
Q.2 In (3), why it is better to declare the $x$-intersept of line in (2) to be next approximate root?

I had learned Newton's method around 12 years back (And not relooked after it), so I didn't get the clear picture of the above process.
 A: Q1: 2 has no significance whatsoever, except that it recreates the sequence from the top of your post. In this case, as long as you pick a positive number, you'll converge to $\sqrt2$. They picked it because it's nice. In general, though, you'd want your first approximate to be a decent guess. And you can't start without picking something.
Q2: Because that's basically the only thing you have to work with. What else should the next approximate root be? Also, as pointed out in the comments below, the whole idea behind Newton's method is to take what we know about the function at a single point (value and derivative), and guess where the function crosses the $x$ axis as well as we can. Assuming that the function looks somewhat like its tangent at that point is a good approximation as long as we're already decently close to the $x$-axis intercept.
A: 
Q.2 In (3), why it is better to declare the x-intercept of line in (2) to be next approximate root?

What we really want is the x-intercept of $f$. Since we're pretending not to know how to solve $f$ directly, we construct a local approximation and find its x-intercept instead.
A: Let $a>0$ and $k\in \mathbb{N}, k>1$. Then $\sqrt[k]{a}$ is the unique positive solution of $x^k-a=0$. Let $f$ be defined by $f(x):=x^k-a$. Since $f'(x)=k x^{k-1}$ the Newton iteration becomes $x_{n+1}=\frac1k((k-1) x_n+\frac{a}{x_n^{k-1}})$. If can be shown that for any $x_0>0$ the sequences $(x_n)_{n\geq 1}$ is monotonically decreasing and convergent with limit equal to $\sqrt[k]{a}$. If seems that this answers your question(s).
A: For the sake of the discussion, let me use
$$b_n:=\frac{a_n}{\sqrt2}$$ so that the recurrence is
$$b_{n+1}=\frac12\left(b_n+\frac1{b_n}\right)=\frac{b_n^2+1}{2b_n}.$$
Now observe that
$$\frac{b_{n+1}-1}{b_{n+1}+1}=\frac{b_n^2-2b_n+1}{b_n^2+2b_n+1}=\left(\frac{b_n-1}{b_n+1}\right)^2$$ so that by induction
$$\frac{b_{n}-1}{b_{n}+1}=\left(\frac{b_0-1}{b_0+1}\right)^{2^n}.$$
As you can check, for any $b_0>0$, $\left|\dfrac{b_0-1}{b_0+1}\right|<1$ and the sequence always converges to zero, so that $b_n$ always converges to $1$, and the convergence is very fast.

In the particular case of the square root, there is a simple explanation of why it works. Still considering the $b_n$, if $b_n>1$, then $\dfrac1{b_n}<1$ and conversely: if the approximation is too large, its inverse is too small. Then the arithmetic mean of the two approximations is closer to the exact root.
