inquality

chrislav

Junior Member
Given A>O find B>0 such that :

For all x>0 and x>B then $|\frac{x}{x-\lfloor x^2\rfloor}|<A$Without using the concept of the limit
Here i dont know how even to start

JeffM

Elite Member
I take it you mean A > 0 rather than A > O.

Is it not obvious that if 0 < B < 1, the problem cannot be solved? If B < 1, then x = 1 is supposedly possible, which entails that the left hand expression is not a real number.

Otis

Elite Member
Interesting piecewise graph.

chrislav

Junior Member
I take it you mean A > 0 rather than A > O.

Is it not obvious that if 0 < B < 1, the problem cannot be solved? If B < 1, then x = 1 is supposedly possible, which entails that the left hand expression is not a real number.
the problem is clear it demands the existence of a B>0 It does not demant the existence of a B between 0 and 1 .If the B is greater than 1 is still gtreater that zero

JeffM

Elite Member
the problem is clear it demands the existence of a B>0 It does not demant the existence of a B between 0 and 1 .If the B is greater than 1 is still gtreater that zero
Yes. But you missed my point entirely.

$0 < B < 1 \implies 1 > B.$
The problem says that the indicated expression is less than some (presumably finite) number for ALL x > B. That is false if x = 1. Therefore, you can work from the assumption that [imath]B \ge 1.[/imath]

JeffM

Elite Member
I am dubious that we have the problem correctly specified.

$0 < x < 1 \implies 0 < x^2 < 1 \implies \lfloor x^2 \rfloor = 0 \implies \dfrac{x}{x - \lfloor x^2 \rfloor} = 1.$
So there can be no B that makes the statement true if A = 0.25.

Please give the wording of the problem completely and exactly.

chrislav

Junior Member
I am dubious that we have the problem correctly specified.

$0 < x < 1 \implies 0 < x^2 < 1 \implies \lfloor x^2 \rfloor = 0 \implies \dfrac{x}{x - \floor x^2 \rfloor} = 1.$
So there can be no B that makes the statement true if A = 0.25.

Please give the wording of the problem completely and exactly.
given A=0.25 put B=6 (B=5 is the lowest value)
then for all x>6 :$|\frac{x}{x-\lfloor x^2\rfloor}|<0.25$

chrislav

Junior Member
I think if you look at the graph of otis then you can easily see that:

Given A>0 THERE always exists a B>0 s.t

fol all x>B ,then : $|\frac{x}{x-\lfloor x^2\rfloor}|<A$of course at x=1 the function is not defined

chrislav

Junior Member
Given A>O find B>0 such that :

For all x>0 and x>B then $|\frac{x}{x-\lfloor x^2\rfloor}|<A$Without using the concept of the limit
Here i dont know how even to start
the x>0 is superfluous since x>B>0 implies x>0