Answer
$$\sqrt 2 \approx 1.414213562$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = {x^2} - 2,{\text{ so }}f'\left( x \right) = 2x \cr
& {\text{Using the Newton's Method }}{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \cr
& {\text{Then evaluating }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right) \cr
& {x_{n + 1}} = {x_n} - \frac{{x_n^2 - 2}}{{2{x_n}}} \cr
& {\text{Taking }}{x_1} = 1 \cr
& {x_1} = 1 \cr
& {x_{1 + 1}} = {x_2} = 1 - \frac{{{{\left( 1 \right)}^2} - 2}}{{2\left( 1 \right)}} = \frac{3}{2} = 1.5\,\,\,\, \cr
& {x_{2 + 1}} = {x_3} = 1.5\, - \frac{{{{\left( {1.5\,} \right)}^2} - 2}}{{2\left( {1.5\,} \right)}} = \frac{{17}}{{12}} \approx 1.416666667 \cr
& {x_{3 + 1}} = {x_4} = 1.416666667 - \frac{{{{\left( {1.416666667} \right)}^2} - 2}}{{2\left( {1.416666667} \right)}} \approx 1.414215686 \cr
& {x_{4 + 1}} = {x_5} = 1.414215686 - \frac{{{{\left( {1.414215686} \right)}^2} - 2}}{{2\left( {1.414215686} \right)}} \approx 1.414213562 \cr
& {x_{5 + 1}} = {x_6} = 1.414213562 - \frac{{{{\left( {1.414213562} \right)}^2} - 2}}{{2\left( {1.414213562} \right)}} \approx 1.414213562 \cr
& {\text{We obtain that }}{x_5} \approx {x_6} \cr
& {\text{Thus}}{\text{,}} \cr
& \sqrt 2 \approx 1.414213562 \cr} $$
