MHB Is infinity / infinity equal to 1?

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The discussion centers on the claim that infinity divided by infinity equals one, using Gamma and Product functions as a basis for the proof. Participants emphasize that infinity is a theoretical concept and not a number, highlighting that "infinity/infinity" is an indeterminate form that can yield various results depending on the limits involved. They point out that treating infinity as a number leads to contradictions and undefined expressions. The distinction between different limits, such as lim(x→∞) (ax/x) equating to 'a', is crucial in understanding the nature of infinity. Overall, the consensus is that infinity divided by infinity cannot be simplistically equated to one.
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Dear Members,
I tried to prove this indeterminate form of infinity / infinity as 1. I could come up a reasonable approach
with Gamma and Product functions. I posted my proof as video in Youtube. Here is the URL for the video.
I would like to receive feedback and challenges on where my approach is going wrong in this proof?

Youtube URL : https://www.youtube.com/watch?v=9W2LclHwQzs

I tried to prove that

infinity / infinity = (infinity ^ 0) = Gamma(1) = ProductPI(0) = 1

Thank you for your help!
Pradyumna
 
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Just a comment, but how your attempt deals with following ($$a > 0$$):

$$\lim_{x \to \infty} \frac{ax}{x} = a \lim_{x \to \infty} \frac{x}{x} = a \cdot 1,$$

and one obtains that $$\frac{\infty}{\infty}$$ can take all positive real values...
 
You can't do $\frac{\infty}{\infty}=\infty^{1-1}=\infty^0$ ... that is, treating infinity as if it were a number; otherwise, you'd get all sorts of weird results.
 
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pradyumna1974 said:
Dear Members,
I tried to prove this indeterminate form of infinity / infinity as 1. I could come up a reasonable approach
with Gamma and Product functions. I posted my proof as video in Youtube. Here is the URL for the video.
I would like to receive feedback and challenges on where my approach is going wrong in this proof?

Youtube URL : https://www.youtube.com/watch?v=9W2LclHwQzs

I tried to prove that

infinity / infinity = (infinity ^ 0) = Gamma(1) = ProductPI(0) = 1

Thank you for your help!
Pradyumna

Keep in mind that "infinity / infinity" is a FORM, not a NUMBER. Don't treat it like a number. "infinity / infinity" is a theoretical condition that never is actually attained. There is a very big difference between "increases without bound" and "we arrived somewhere finite".
 
It is possible to formalize the notion of $\infty$ as an "ordinary number" by introducing the "extended real number line" $\overline{\mathbb{R}}$ and topologizing it with the order topology. In my limited experience, this is more of a convenience in certain areas such as real and convex analysis, and not so much a source of insight. Also, the algebraic properties of $\overline{\mathbb{R}}$ are not very good. In particular, expressions such as $\frac{\infty}{\infty}$ are still undefined, for reasons such as those pointed out nicely in post #2 and the other posts above.
 
You use the phrase "indeterminant form". Surely you understand that the reason that $\frac{\infty}{\infty}$ is called an "indeterminant form" is that taking the limit as the numerator and denominators go to infinity, in different ways will give different results. Certainly $\lim_{x\to\infty} \frac{x}{x}= 1$ but $\lim_{x\to\infty} \frac{ax}{x}= a$ for any value of a. all of which can be thought of as "$\frac{\infty}{\infty}$".
 
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There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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