Why is 1^∞ an Indeterminate Form?

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Discussion Overview

The discussion centers around the mathematical concept of the indeterminate form \(1^\infty\). Participants explore why this form is considered indeterminate, examining various limits and the behavior of functions approaching this form. The scope includes mathematical reasoning and limit evaluation.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions why \(1^\infty\) is considered indeterminate, suggesting that multiplying 1 by itself should always yield 1.
  • Another participant proposes that \(1^\infty\) is indeterminate because it relates to the form \(0 \cdot \infty\) and mentions taking the logarithm of \(1^\infty\) as a mnemonic trick.
  • Several participants provide examples of limits that approach the form \(1^\infty\), such as \(\lim_{x\rightarrow \infty} (1+\frac{1}{x})^{e^{x^5}}\), noting that while \(1+\frac{1}{x}\) approaches 1, it does so at a different rate compared to \(e^{x^5}\) approaching infinity.
  • Another example given is \(\lim_{x\rightarrow \infty} (\cos(1/x^2))^{x^4}\), suggesting that limits of the form \(1^\infty\) can yield different values depending on the functions involved.
  • A participant expresses uncertainty about the concept of speed of convergence and refrains from providing a detailed explanation, referencing L'Hospital's rule as a related concept.

Areas of Agreement / Disagreement

Participants do not reach a consensus on why \(1^\infty\) is indeterminate, with multiple competing views and examples presented. The discussion remains unresolved regarding the underlying reasons for the indeterminate nature of this form.

Contextual Notes

Participants express uncertainty about the concept of speed of convergence and its implications for limits approaching \(1^\infty\). There are indications of missing assumptions and the complexity of the topic, which may affect the clarity of the discussion.

mbrmbrg
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Why is 1^\infty an indeterminate form? If you keep multiplying 1 by itself doesn't the answer stay 1?
 
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It is an indeterminate form because 0 . \infty is.

Try to take the "logarithm of 1^\infty".

EDIT:
Obviously it is not a proof.
It is only a mnemonic trick.
 
Last edited:
Sure,

\lim_{x\rightarrow \infty} 1^x=1.

But consider a case like

\lim_{x\rightarrow \infty} (1+\frac{1}{x})^{e^{x^5}}

1+1/x goes to 1, but does it so converges much slower than exp(x^5) goes to infinity! So if you think about it this way, maybe you see that it makes sense that sometimes a limit of the "form" 1^{\infty}[/tex] will diverge, and some other times, it will behave and go to 1.<br /> <br /> Maybe in some cases, the limit will converge to a value that is not 1? Can someone provide an exemple?
 
Last edited:
quasar987 said:
Sure,

\lim_{x\rightarrow \infty} 1^x=1.

And, obviously,

\lim_{x\rightarrow \infty} 0 x = 0


Maybe in some cases, the limit will converge to a value that is not 1? Can someone provide an exemple?

\lim_{x\rightarrow \infty} (\cos(1/x^2))^{x^4}
 
Last edited:
quasar987 said:
Sure,

\lim_{x\rightarrow \infty} 1^x=1.

But consider a case like

\lim_{x\rightarrow \infty} (1+\frac{1}{x})^{e^{x^5}}

1+1/x goes to 1, but does it so converges much slower than exp(x^5) goes to infinity! So if you think about it this way, maybe you see that it makes sense that sometimes a limit of the "form" 1^{\infty}[/tex] will diverge, and some other times, it will behave and go to 1.<br /> <br /> Maybe in some cases, the limit will converge to a value that is not 1? Can someone provide an exemple?
<br /> <br /> Yeah, that&#039;s what I was talking about. I know that the form 1^{\infty} has many different values; I wanted to know <i>why</i>.<br /> Can you run through the &quot;converging at different speeds&quot; bit a lot slower (pun unintended, but noted and pride taken)?
 
I started writing an explanation of the concept of speed convergence, but in the middle of it, I realized that I'm too unsure about too many points, and the explanation is unnecessarily too blurry imo. So I'll let someone more qualified answer.

Informally, the concept of speed convergence is incarnated in L'Hospital's rule: if both f and g go to infinity, then the limit of their ratio is determined by the ratio of their speed (derivative) at infinity.
 
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