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Indeterminate forms

  1. Aug 26, 2009 #1
    I'm just curious, why, when solving limits, is [itex]1^\infty[/itex] considered an indeterminate form? Isn't 1 raised to any power equal to 1? Why isn't it so simple?
     
  2. jcsd
  3. Aug 26, 2009 #2
    Probably because if you perturb the 1 at all, the result is either zero or infinite.
     
  4. Aug 26, 2009 #3
    Well that is a very good question but the problem is that it depends on how fast your function is going to 1 or infinity.

    Your function can be going so slowly to 1, in which case the limit goes to 0.

    It`s the same as the undetermined form 0/0 the function on top and bottom could approach zero at the same speed and the limit could go to 1.
     
  5. Aug 26, 2009 #4
    Ah, yes, the function never actually reaches the "value" [itex]1^\infty[/itex], since we never consider the function at exatcly [itex]x=a[/itex], we are just curious about what it does around that point.

    Thank you for your answers.
     
    Last edited: Aug 26, 2009
  6. Aug 26, 2009 #5
    Precisely, you are correct. Even in the precise definition of a limit we only look at the deleted neighborhood of x .

    You are welcome.:)
     
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