# How do you get x_n>|x|/2 from 0<|x_n - x|<|x|/2 ?

Oct 21, 2017

See below.

#### Explanation:

From $0 < | {x}_{n} - x | < | x \frac{|}{2}$ we have equivalently

${\epsilon}_{1}^{2} = \sqrt{{\left({x}_{n} - x\right)}^{2}}$
$\sqrt{{\left({x}_{n} - x\right)}^{2}} + {\epsilon}_{2}^{2} = \sqrt{{\left(\frac{x}{2}\right)}^{2}}$

with $\left\{{\epsilon}_{1} , {\epsilon}_{2}\right\} \ne 0$

then

${\left(\sqrt{{\left({x}_{n} - x\right)}^{2}} + {\epsilon}_{2}^{2}\right)}^{2} = {\left(\frac{x}{2}\right)}^{2}$ or

${\left({x}_{n} - x\right)}^{2} + 2 {\epsilon}_{2}^{2} \sqrt{{\left({x}_{n} - x\right)}^{2}} + {\epsilon}_{2}^{4} = {x}^{2} / 4$

and again

${\left(2 {\epsilon}_{2}^{2} \sqrt{{\left({x}_{n} - x\right)}^{2}}\right)}^{2} = {\left({x}^{2} / 4 - {\left({x}_{n} - x\right)}^{2} - {\epsilon}_{2}^{4}\right)}^{2}$

or factoring

-1/16 (2 epsilon_2^2 + x - 2 x_n) (2 epsilon_2^2 + 3 x - 2 x_n) (2 epsilon_2^2 - 3 x + 2 x_n) (2 epsilon_2^2 - x + 2 x_n)=0

or

{(2 epsilon_2^2 + x - 2 x_n=0), (2 epsilon_2^2 + 3 x - 2 x_n=0), (2 epsilon_2^2 - 3 x + 2 x_n=0), (2 epsilon_2^2 - x + 2 x_n=0):}

or equivalently

{(x/2 - x_n < 0), (3/2 x - x_n < 0), ( - 3/2 x + x_n < 0), (- x/2 + x_n < 0):}

or equivalently

$\frac{x}{2} < {x}_{n}$ and ${x}_{n} < \frac{x}{2}$

$\frac{3}{2} x < {x}_{n}$ and ${x}_{n} < \frac{3}{2} x$