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Flofloflo
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Joined: 07 Nov 2007 Posts: 120

Posted: 30 Aug 2009 08:13:21 am Post subject: 


Hello,
I was wondering about infinetely small numbers. 0,999... equals one, according to my teachers, and also according to me.
But then what would you get if you take lim n→∞ int(11/n)?
Would that equal 1 because lim n→∞ 1/n = 0, or would it equal 0 because 1 1/n is always slightly smaller than 1?? 

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thornahawk μολών λαβέ
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Joined: 27 Mar 2005 Posts: 569

Posted: 30 Aug 2009 08:21:47 am Post subject: 


Flofloflo wrote: ...Would that equal 1 because lim_n→∞ 1/n = 0...
Yes. :)
Alternatively, interpret 0.999... as an infinite geometric series.
thornahawk 

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FloppusMaximus
Advanced Member
Joined: 22 Aug 2008 Posts: 472

Posted: 30 Aug 2009 11:36:47 pm Post subject: 


I assume by 'int' you mean the floor (or integerpart) function. Then yes, lim_{x→∞} ⌊1  1/x⌋ = 0. (Whereas lim_{x→∞} (1  1/x) is equivalent to 0.999..., which equals 1.) 

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DarkerLine ceci n'est pas une 
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Joined: 04 Nov 2003 Posts: 8328

Posted: 31 Aug 2009 12:16:09 am Post subject: 


By the way, the reason you can't substitute 1/n→0 in that limit is that ⌊x⌋ is not continuous. In fact, it has a discontinuity at precisely the right point (x=1) to ensure that lim_{n→∞} ⌊11/n⌋ is not the same as ⌊lim_{n→∞} 11/n⌋.
the sequence ⌊11/n⌋ is always 0, so its limit is also 0.
Last edited by Guest on 31 Aug 2009 12:20:42 am; edited 1 time in total 

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Flofloflo
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Joined: 07 Nov 2007 Posts: 120

Posted: 31 Aug 2009 04:31:52 pm Post subject: 


Wait.... Maybe I am misunderstanding stuff but it seems you disagree with Tornahawk???
With Int I mean the 'Entier function' which means: substract everything behind the decimal place (3,5 becomes 3 etc.).
But basically what Darkerline says implies that there is such a thing as a 'smallest number larger then zero', because this number has the characteristic that it can change the entier function of an integer. 

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darkstone knight
Advanced Member
Joined: 07 Sep 2008 Posts: 438

Posted: 31 Aug 2009 04:40:33 pm Post subject: 


u=1/n
u(n) is a multiple of u(n1)
thus if u(n) = 0
u(n1) must be 0 aswell
thus, u(1) must be 0, wich is not the case
:confused: 

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simplethinker snjwffl
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Joined: 25 Jul 2006 Posts: 700

Posted: 31 Aug 2009 05:01:55 pm Post subject: 


Flofloflo wrote: Wait.... Maybe I am misunderstanding stuff but it seems you disagree with Tornahawk???
With Int I mean the 'Entier function' which means: substract everything behind the decimal place (3,5 becomes 3 etc.).
lim_{n→∞}(1  1/n) and lim_{n→∞}( int(1  1/n) ) are two totally different things. In the first sequence (with 11/n), the first few values (starting with n=1, since n=0 makes them explode) are {0, 1/2, 2/3, 3/4, 4/5, ...}. For the second sequence, it takes the values {int(0), int(1/2), int(2/3), int(3/4), int(4/5), ...} = {0, 0, 0, 0, 0, ...}. That's why it looks like DarkerLine is disagreeing with Thornahawk since they're talking about two different limits.
Quote: But basically what Darkerline says implies that there is such a thing as a 'smallest number larger then zero', because this number has the characteristic that it can change the entier function of an integer.
Depending on what set of numbers you're using there is a smallest number larger than zero. In the integers or natural numbers "1" is the smallest number larger than zero (and applying "int" or "floor" to the sequence is essentially restricting the values to the integers).
Last edited by Guest on 05 Jul 2010 08:09:53 am; edited 1 time in total 

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DarkerLine ceci n'est pas une 
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Joined: 04 Nov 2003 Posts: 8328

Posted: 31 Aug 2009 07:30:28 pm Post subject: 


Flofloflo wrote: But basically what Darkerline says implies that there is such a thing as a 'smallest number larger then zero', because this number has the characteristic that it can change the entier function of an integer. What I'm saying implies no such thing. The problem is that int() doesn't behave in the way you intuitively expect limits to work, because it's not continuous.
I mean, if you wanted to, you could talk about hyperreal numbers here. In that case, you wouldn't really be talking about limits, and instead of taking a limit, you would plug in an infinite number to int(11/n), and possibly take the standard part of the result. But the usual way of extending a real function such as int() to the hyperreals would still not get you the sort of number you're looking for, because if 0<x<1, then int(x)=0 even for a hyperreal x.
Last edited by Guest on 31 Aug 2009 07:42:38 pm; edited 1 time in total 

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FloppusMaximus
Advanced Member
Joined: 22 Aug 2008 Posts: 472

Posted: 31 Aug 2009 10:32:25 pm Post subject: 


darkstone knight: I have no idea what you're trying to say, but you probably divided by zero or something.
Flofloflo: Keep a couple of things in mind:
(1) Infinity is not a number, so x cannot equal infinity.
(2) The limit of f(x) as x approaches a has nothing whatsoever to do with the value of f(a).
When we talk about "the limit of f(x) as x approaches infinity", we mean a number L, such that as x becomes very large, f(x) becomes very close to L. It doesn't matter whether f(x) ever equals L. But in the case of f(x) = ⌊1  1/x⌋, that function is always zero for x > 1, so it doesn't make sense for the limit to be anything other than zero. In fact, its limit is zero: for any ε > 0, and for any x > 1, f(x)  0 < ε. 

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