Question about indeterminate form

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Discussion Overview

The discussion revolves around the concept of indeterminate forms in calculus, specifically focusing on the form $$1^\infty$$ and its implications in limit calculations. Participants explore various examples and definitions related to indeterminate forms, including other expressions involving infinity.

Discussion Character

  • Debate/contested

Main Points Raised

  • Some participants assert that $$1^\infty$$ is an indeterminate form because limits can yield values other than 1, citing the example of $$\lim_{x \to \infty}(1 + \frac{1}{x})^x$$ which approaches e.
  • Others argue that anything involving $$\infty$$ is an indeterminate form, as $$\infty$$ is not a number.
  • There is a claim that $$1/\infty$$ is not an indeterminate form because its value is definitively 0.
  • Some participants agree that certain forms involving infinity, such as $$[\infty + \infty]$$ and $$[\infty * \infty]$$, are not considered indeterminate, while $$[\infty - \infty]$$ and $$[\frac{\infty}{\infty}]$$ are.
  • Participants discuss whether $$\frac{0}{\infty}$$ is also an indeterminate form, with mixed responses regarding its status.

Areas of Agreement / Disagreement

Participants express disagreement on the classification of various forms involving infinity, particularly $$1^\infty$$ and $$1/\infty$$. There is no consensus on whether $$\frac{0}{\infty}$$ is indeterminate, with differing opinions presented.

Contextual Notes

Limitations in the discussion include varying definitions of indeterminate forms and the reliance on specific examples to illustrate points, which may not encompass all cases.

Maged Saeed
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why $$1^\infty$$ is indeterminate form?
 
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Maged Saeed said:
why $$1^\infty$$ is indeterminate form?
1 raised to any finite power is 1, of course, but some limits are of this indeterminate form, and have a limit that isn't equal to 1.

The most famous example is this limit:
$$\lim_{x \to \infty}(1 + \frac{1}{x})^x$$

The base is approaching 1 and the exponent is "approaching" infinity. It can be shown that the value of this limit expression is the number e.
 
Thanks ,, I got it now
(:
 
Maged Saeed said:
why $$1^\infty$$ is indeterminate form?

Anything with ##\infty## is an indeterminate form, because ##\infty## is not a number.
 
PeroK said:
Anything with ##\infty## is an indeterminate form, because ##\infty## is not a number.

I don't believe that [itex]1/\infty[/itex] is an indeterminate form because its value cannot be anything other than 0.
 
Mentallic said:
I don't believe that [itex]1/\infty[/itex] is an indeterminate form because its value cannot be anything other than 0.
I agree. It's not indeterminate because you can determine what the limit will be.
 
PeroK said:
Anything with ##\infty## is an indeterminate form, because ##\infty## is not a number.
That's not true. While ##[\infty - \infty]## and ##[\frac{\infty}{\infty}]## are indeterminate forms, ##[\infty + \infty]## and ##[\infty * \infty]## are not considered indeterminate.
 
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Mark44 said:
That's not true. While ##[\infty - \infty]## and ##[\frac{\infty}{\infty}]## are indeterminate forms, ##[\infty + \infty]## and ##[\infty * \infty]## are not considered indeterminate.

Mark44 said:
I agree. It's not indeterminate because you can determine what the limit will be.

Me too .

How about
$$\frac{0}{\infty}$$

Is it indeterminate too?

I think so

:oldeyes:
 
Maged Saeed said:
Me too .

How about
$$\frac{0}{\infty}$$

Is it indeterminate too?

I think so

:oldeyes:
No, it is not indeterminate. If the numerator approaches 0 and the denominator becomes unbounded, the limit is 0.
 
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Oh, I see,,
Thanks
 

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