wrong result in calculating limit of x*floor(1/x) as x goes to 0
wrong result in calculating limit of x*floor(1/x) as x goes to 0
wrong result in calculating limit of x*floor(1/x) as x goes to 0
I get the noun form back. Not wrong, but could be better. What were you expecting?
I got the wrong result, namely 0. I attached my file
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
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?
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
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.
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
Log in to post a comment.