#1804 limit of x*floor(1/x) as x goes to 0

None
open
nobody
5
2013-07-26
2009-10-29
Rudolf Vyborny
No

wrong result in calculating limit of x*floor(1/x) as x goes to 0

Discussion

• Rudolf Vyborny
2009-10-29

wrong result in calculating limit of x*floor(1/x) as x goes to 0

Attachments

• Raymond Toy
2009-10-29

I get the noun form back. Not wrong, but could be better. What were you expecting?

• Rudolf Vyborny
2009-10-29

I got the wrong result, namely 0. I attached my file

• Dieter Kaiser
2009-11-09

With Maxima 5.19post I get a noun form too. There have been serveal changes the last time to improve limit. Furthermore, I think the limit of the example is not defined. Therefore, it seems to be not wrong to return a noun form.

Setting the status to pending and works for me.
Dieter Kaiser

• Dieter Kaiser
2009-11-09

• status: open --> pending

• Raymond Toy
2009-11-09

Isn't the limit 1? Let any x small enough, 1/x = n + e, where n is an integer and e < 1. Then floor(1/x) = n and x*floor(1/x) is n/(n+e) = 1 - e/(n+e). As n gets larger (and x gets smaller), this approaches 1.

Did I make a mistake?

• Dieter Kaiser
2009-11-09

Sorry, I have no mathematical proof. I have come to the conclusion because of
the follwing:

1. The function floor(x) is discontinuous.
2. The function x*floor(1/x) has an infinite number of points of discontinuity
in any infinitesimal intervall when aproching zero.
3. Therefore, the function does not approach a limit.

I could be wrong.

Dieter Kaiser

• Rudolf Vyborny
2009-11-11

Answer to rtoy. The limit is 1, your proof is essentially orrect. I prefer the following proof: 1/x-1<floor(1/x)<=1/x, hence for
x>0 we have 1-x<xfloor(1/x)<=1. It follows that the limit from the right is 1. The proof for the limit from the left is similar.
Answer to crategus: The function does have a limit, namely 0. Your staement 2. is correct but your conclusion 3. is erroneous.

• Rudolf Vyborny
2009-11-11

• status: pending --> open

• Dieter Kaiser
2009-11-12

For the record: Maxima 5.19post gives the result:

(%i2) limit(x*floor(1/x),x,0);
(%o2) 'limit(floor(1/x)*x,x,0)

That is we get noun form. We expect the answer 1.

Changing the title of this bug report to reflect the problem better and the resolution ID to none.

Dieter Kaiser

• Dieter Kaiser
2009-11-12

• summary: wrong result --> limit of x*floor(1/x) as x goes to 0

• Robert Dodier
2013-07-26

• labels: --> Lisp Core - Limit
• Group: --> None

• Robert Dodier
2013-07-26