inequality 2

[chrislav]

Given A>0 find a B>0 such that:
For all x>0, x>B
$$x\neq 1$$ then $$\frac{x}{x-\lfloor x\rfloor}>A$$without using the concept of the limit

 

[blamocur]

This somewhat suspicious: whatever the value of B you can find an integer $x > B$ -- for example $\lceil B\rceil+1$ -- which means that $x-\lfloor x\rfloor = 0$, and the fraction becomes undefined (or infinity if you wish).

 

[JeffM]

Where is this expression defined!

 

[chrislav]

This somewhat suspicious: whatever the value of B you can find an integer $x > B$ -- for example $\lceil B\rceil+1$ -- which means that $x-\lfloor x\rfloor = 0$, and the fraction becomes undefined (or infinity if you wish).

YES i checked with Alpha Wolfram and the limit is infinity

 

[blamocur]

If you don't care about zero in the denominator then my hint is: start with an inequality for that denominator, i.e., $x-\lfloor x\rfloor$.

 

[chrislav]

Where is this expression defined!

all the real Nos except inegers