Inequality

[solakis1]

given $A>0$ find a $B>0$ such that:

For all $x>0$ and $x>B$ Then $|\frac{x}{x-\lfloor x^2\rfloor}|<A$
Do not use the concept of the limit

 

[solakis1]

hint:[sp] $|\frac{x}{x-\lfloor x^2\rfloor}|$=$|\frac{1}{x}|.|\frac{1}{\frac{1}{x}-\frac{\lfloor x^2\rfloor}{x^2}}|$[/sp]