Question

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

Answer 1

See below.

Explanation:

From `0<|x_n - x|<|x|/2` we have equivalently

`epsilon_1^2=sqrt((x_n-x)^2)`
`sqrt((x_n-x)^2)+epsilon_2^2 = sqrt(((x)/2)^2)`

with `{epsilon_1,epsilon_2} ne 0`

then

`(sqrt((x_n-x)^2)+epsilon_2^2)^2 = ((x)/2)^2` or

`(x_n-x)^2+2epsilon_2^2 sqrt((x_n-x)^2)+epsilon_2^4=x^2/4`

and again

`(2epsilon_2^2 sqrt((x_n-x)^2))^2=(x^2/4-(x_n-x)^2-epsilon_2^4)^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

`x/2< x_n` and `x_n < x/2`

`3/2x < x_n` and `x_n < 3/2 x`

Those results are contradictory.