If we approximate $\sqrt{2}$ using Newton's method for $f(x)=x^2-2$, we get a convergent sequence $\{x_n\}_{n\in\mathbb{N}}$ that satisfies the following recurrence formula:
$$x_{n+1}=\frac{1}{2}\left(x_n+\frac{2}{x_n}\right)$$
It it easy to prove that the error $e_n:=x^n-\sqrt{2}$ satisfies $\frac{e_{n+1}}{e_n}<\frac{1}{2}$, so it is roughly better than the bisection method.
I would like to know the value or evaluation of the following limit:
$$\varlimsup_{n\rightarrow \infty}e_n2^n$$
When $x_0>\sqrt{2}$ we have $e_0>0$ and $$e_{n+1}=x_{n+1}-\sqrt{2}={x_n^2-2x_n\sqrt{2}+2\over 2x_n}={\bigl(x_n-\sqrt{2}\bigr)^2\over2x_n}={e_n^2\over 2x_n}>0\qquad(n\geq0)\ .$$ This shows that the convergence of Newton's method is quadratic. Hence this method is not only "roughly better" than the bisection method.
For the second question put $a_n:=2^ne_n>0$. We then have $$a_{n+1}=2^{n+1} {e_n^2\over 2x_n}=a_n{e_n\over x_n}\ .$$ Since $e_n\to0$ and $x_n\to\sqrt{2}$ when $n\to\infty$ there is some $n_0$ such that $$a_{n+1}\leq{1\over2}a_n\qquad(n>n_0)\ .$$ This proves $\lim_{n\to\infty} a_n=0$.