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Zero times infinity 
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#1
Apr2706, 04:11 AM

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0 * infinity
What is up? I read that it this expression is called an "indeterminate form." Why isn't zero multiplied by infinity equal to zero? Even if infinity is really big, if there are zero amounts of infinity, that would make zero. How about 1^{infinity}? 1? Guess not. infinity / infinity? 1? Guess not. 


#2
Apr2706, 04:17 AM

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Infinity, not being part of the natural, rational, real or complex numbers, does not come with the innate possibility to multiply it by anything in those sets. We can define a consistent notion of arithmetic on the extended numbers (gotten by adding in a symbol for infinity) in many cases. To be consistent with these definitions then we cannot define the multiplication 0*infinity. It is no big deal, and nothing to do with 'zero amounts of anything'. You must not think of the mathematical notion of infinity as being the same as your 'really big number, bigger than you can imagine' unmathematical notion.



#3
Apr2706, 04:53 AM

P: 3,243

my book states that in some cases you can calculate 0*infinity by using l'hopital rule when 0*[tex]\infty[/tex]=(1/[tex]\infty[/tex])*[tex]\infty[/tex].



#4
Apr2706, 05:07 AM

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Zero times infinity
that is analysis, that is about taking limits of things *that tend to infinity*. It is lazy short hand to actually say it 'is' a statement about infinity, rather than what it actually is: a statement about limits.
As a demonstration, suppose g(x)=0 at x=0 and suppose f(x) tends to infinity as x tends to zero (lazy short hand would be f(0)=infinity). Then f(x)g(x), as x went to zero, if it were to exist, would be '0*infinity' in the lazy short hand. Picking f(x)=1/x and g(x)=nx for any real number n, then the result is n, hence indeterminate. 


#5
Apr2706, 07:16 AM

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#6
Apr2706, 07:22 AM

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So, does
[tex] e = \lim_{x \rightarrow \infty} \left(1 + \frac{1}{x} \right)^x = \left( \lim_{x \rightarrow \infty} 1 + \frac{1}{x} \right)^ {\lim_{x \rightarrow \infty} x} = 1^\infty = 1 [/tex] ? 


#7
Apr2706, 08:28 AM

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#8
Apr2706, 08:34 AM

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Start multiplying them, and I'll ask you in a few hours if you have reached an answer. Your answer, of course, will be "the answer has not been determined yet, as I am not finished the calculation." And then I'll come back in a few months and ask again. What do you suppose your answer will be then? Remember, there is no guessing or anticipating in math... 


#9
Apr2706, 10:17 AM

P: 2,046

Thanks for all the comments, guys.
Btw, how does one resolve the apparent contradiction in the Hurkyl's post? That's the first time that I've seen a limit applied to the power. 


#10
Apr2706, 10:44 AM

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What contradiction in Hurkyl's post? There is no reason to suppose the limits behave like that, and that demonstrates that they do not. Just because you can manipulate some symbols does not mean you may do so and have the result make sense.



#11
Apr2706, 01:53 PM

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Hurkyl's post is a reductio ad absurdem. It is a proof that
[tex]\lim (a\circ b)\neq \lim a \circ \lim b[/tex] where o is some operation. 


#12
Apr2706, 06:06 PM

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The contradiction is resolved by discarding the hypothesis that [itex]1^{\infty}[/itex] is a determinate form. As krab said, it's a proof by contradiction. If [itex]1^{\infty}[/itex] is determinate, then my manipulations are justified. But, my manipulations derive a contradiction, and thus [itex]1^{\infty}[/itex] is indeterminate. (Those should all be positive infinities, but it doesn't typeset well) 


#13
May906, 11:16 PM

P: 2

It is true that the actual number infinity is unlimited, however the only way to perform any mathematical calculation is one step at a time. But imagine one^infinity as a line, as a graph of y=1^x. That is, of course, a straight line. Since said line will continue forever, we can see that it can "reach" infinity, although infinity can never be reached in the real sense. Basically, it will go on forever, so 1^infinity =1



#14
May1006, 12:01 AM

P: 212

Consider the function f(x) = (1+x)^(1/x)
Clearly the "limit" of this function as x approaches 0 is 1^infinity, yet try to graph it. The actual limit is e, 2.71828, which is then "equal" to 1^infinity. Clearly e is not equal to 1 so therefore 1^infinity doesn't always have to equal 1 (i.e., it's indeterminate) The "one step at a time" thing is irrelevant; when dealing with infinities, you have to accept that there can be "supertasks", i.e. tasks that are doable but unable to complete in a finite series of steps. 


#15
May1006, 12:50 AM

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I guess what they are saying is that the limit of a^b, where a approaches 1 and b approaches infinity, is hard to determine.
this should not be too shocking, since if we take the log, we are back to matts exampels of how the limit of a product can be anything, when one factor approaches zero and the other approaches infinity. for instance it is not hard to calculate the limit of log(1 + 1/x) as x goes to infinity from the integral definition of log, and estimate it as between 1/x+1 and 1/x. hence x times it goes to 1, which gives the limit of xlog(1 + 1/x), i.e. the log of (1 + 1/x)^x. since the log goes to 1, the original limit is e. 


#16
May1006, 03:15 AM

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#17
Jan3008, 04:00 PM

P: 2

[tex] \lim_{x \rightarrow \infty} 1^x =1 [/tex] Certainly you can take the next step and define [tex]1^{\infty}[/tex] to be one if you like. But this definition is only one of many inconsistent definitions that could be made. Since [tex] \lim_{x \rightarrow \infty}{\left(1+\frac{1}{x}\right)^{x}} = e [/tex] why not define [tex]1^{\infty}[/tex] as e? Or for that matter, as any arbitrarily chosen number, since: [tex] \lim_{x \rightarrow \infty} \left(1+\frac{1}{x}\right)^{n x} = e^{n} [/tex] The crux of the matter is, infinity doesn't obey the rules that numbers obey. It can't  this is provable. A great deal of trouble occurs when people take laws of arithmetic and try to apply them to infinity. If you assume that infinity obeys all the same laws of arithmetic, then you have assumed a contradiction, and are able to "prove" anything. [tex]\lim_{m\rightarrow\infty}\lim_{n\rightarrow\infty}\frac{m}{m+n}[/tex] The answer you get depends on the order in which you take the limits. You have to be very careful when working with limits to make sure each step in a derivation is justified. The first step in Hurkyl's isn't. 


#18
Feb108, 04:58 AM

P: 61

not_e = \lim_{x \rightarrow \infty} \left(1 \right)^x = \left( \lim_{x \rightarrow \infty} 1 \right)^ {\lim_{x \rightarrow \infty} x} = 1^\infty = 1 [/tex]? 


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