inquality

chrislav

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Given A>O find B>0 such that :

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

JeffM

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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.
 

chrislav

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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

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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.

[math]0 < B < 1 \implies 1 > B.[/math]
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

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I am dubious that we have the problem correctly specified.

[math]0 < x < 1 \implies 0 < x^2 < 1 \implies \lfloor x^2 \rfloor = 0 \implies \dfrac{x}{x - \lfloor x^2 \rfloor} = 1.[/math]
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

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I am dubious that we have the problem correctly specified.

[math]0 < x < 1 \implies 0 < x^2 < 1 \implies \lfloor x^2 \rfloor = 0 \implies \dfrac{x}{x - \floor x^2 \rfloor} = 1.[/math]
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 :[math]|\frac{x}{x-\lfloor x^2\rfloor}|<0.25[/math]
 

chrislav

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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 : [math]|\frac{x}{x-\lfloor x^2\rfloor}|<A[/math]of course at x=1 the function is not defined
 

chrislav

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Given A>O find B>0 such that :

For all x>0 and x>B then [math]|\frac{x}{x-\lfloor x^2\rfloor}|<A[/math]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
 
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