MHB Can a Constant be Chosen to Satisfy an Inequality for All Real Numbers?

[solakis1]

Prove or disprove the following:
There exists $A$ such that for all $a>0$ there exists $b>0$ such that for all $ x$:

$|x-\ x_0|<b$ i mplies. $|\frac{1}{[x]}-A|<a$ where [x] is the floor value of x

Gvf

 

[Opalg]

Prove or disprove the following:
There exists $A$ such that for all $a>0$ there exists $b>0$ such that for all $ x$:

$|x-\ x_0|<b$ i mplies. $|\frac{1}{[x]}-A|<a$ where [x] is the floor value of x

How does $x_0$ come into this?

 

[solakis1]

put A= $\frac{1}{[x_0]}$ and $[x_0]$ is not 0

 

[Opalg]

You have not explained the status of $x_0$. Is it a fixed number given in advance, or are we supposed to prove that the result holds for all $x_0$?

 

[solakis1]

With $x_0$ we usualy define a constant that can take several different values but not integers for example 2.3, 4.1,12.9 e.t.c. e. t. c