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For me, multiplication is a binary operation so it can be applied only on a finite sequence of numbers. but $1^{\infty}$ requires that we apply multiplication infinitly which is not defined as multiplication is a binary operation.

Is that a good reason? If not, what is the reason?

If my reason is ok, So similarly, $5^{\infty}$ is indeterminate ?

ِAdded:

I noticed that all answers are in context of "limits". Algebraically, multiplication is a binary operation, So it ONLY can be used to define multiplication of finite sequence of numbers as not a infinite sequence. So algebraically, what does $1^{\infty}$ even mean?

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    $\begingroup$ E.g., compare $1^n$ and $(1+1/n)^n$, for large $n$. $\endgroup$ – David Mitra Oct 5 '14 at 9:16
  • $\begingroup$ Let $A=1^{\infty}, \ln A=\infty\ln1=\dfrac00$ Similarly, $B=5^{\infty}, \ln b=\infty\ln5=\infty$ $\endgroup$ – lab bhattacharjee Oct 5 '14 at 9:16
  • $\begingroup$ @RainiervanEs , My question is different! I'm asking from an algebraic point of view as binary operation see the added part not in context of limits. $\endgroup$ – Fawzy Hegab Oct 5 '14 at 9:23
  • $\begingroup$ @MathsLover: Any time you mention $\infty$, you are talking about limits. Without speaking of convergence, $1^\infty$ is unlikely to mean anything. $\endgroup$ – Eric Stucky Oct 5 '14 at 9:23
  • $\begingroup$ @labbhattacharjee,$ln(a^n)=n ln(a)$ only true when $n$ is real number not infinity. $\endgroup$ – Fawzy Hegab Oct 5 '14 at 9:24
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The reason on why this is indeterminate is because of how it behaves when we go to limits. For example, if you look at $1^n$ as $n\rightarrow\infty$ we would get that $1^\infty=1$, while by looking at $(1+\frac{1}{n})^n$ we get that $1^\infty=e$. We don't get such thing with $5^\infty$ because, no matter what, it always diverges to infinity.

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  • $\begingroup$ Why not to define $1^\infty$ as $lim_{x\rightarrow \infty} 1^x$ neglecting the other limit? Why should we consider the other limit at all? I knowthat the two expressions get closer to each other as $x$ goes larger but in any given real $n$ ,they still are not the same! $\endgroup$ – Fawzy Hegab Oct 5 '14 at 10:03
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    $\begingroup$ You can define it like that, but this is one of infinitely many possible definitions. Remember that for determinate forms, like $2^4$, the result of $\lim_{x\rightarrow 2,y\rightarrow 4}x^y$ does not depend on the way that $x,y$ tend to their limits. $\endgroup$ – Wojowu Oct 5 '14 at 10:06
  • $\begingroup$ So, Can we say that we define $a^n$ to be the limit $lim_{x\rightarrow a,y\rightarrow n}x^y$? So if the limit exists then it's a determinate form otherwise indetermindate form? Is that the standard definition? $\endgroup$ – Fawzy Hegab Oct 5 '14 at 10:32
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    $\begingroup$ I believe this is a definition most people would agree with. $\endgroup$ – Wojowu Oct 5 '14 at 10:36
  • $\begingroup$ But according to that definition, $5^\infty$ is indeterminate as in the link: wolframalpha.com/input/… $\endgroup$ – Fawzy Hegab Oct 5 '14 at 11:00
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Because $1=a^0$, and $0\cdot\infty$ is $($also$)$ undetermined.

Because all convergent infinite products are of the form $1^\infty$, since their general term tends to $1$, and the number of terms is infinite, but they don't all converge to the same value. Furthermore, there are also divergent infinite products whose general term also tends to $1$.

Etc.

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  • $\begingroup$ I don't uderstand the paragraph "because all .... ", could you clarify plz? $\endgroup$ – Fawzy Hegab Oct 5 '14 at 9:43
  • $\begingroup$ @MathsLover: Were you taught infinite products ? $\endgroup$ – Lucian Oct 5 '14 at 9:48
  • $\begingroup$ I never knew that $1^\infty$ can be interpreted as infinite product. It's quite unnatural for me, because some of the terms in such product can be quite far from 1. $\endgroup$ – Wojowu Oct 5 '14 at 9:57
  • $\begingroup$ @Lucian , No but Whenever I was seing thing like $a^\{infty}$ i was interpreting it according to this definition. $\endgroup$ – Fawzy Hegab Oct 5 '14 at 10:09
  • $\begingroup$ @Wojowu: Infinity minus some is still infinity. The same reasoning applies to infinite sums, where the general term tends to $0$. Though for $0\cdot\infty$ we have simpler ways of showing it to be undetermined. $\endgroup$ – Lucian Oct 5 '14 at 10:20

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