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Indeterminate form - 1^∞

  1. May 10, 2012 #1
    I am sure that [itex]x^{\infty}[/itex] as [itex]x\to1[/itex] is an indeterminate form.

    But can someone please explain how [itex]1^{\infty}[/itex] is indeterminate? I always thought it is equal to 1.
  2. jcsd
  3. May 10, 2012 #2
    We choose [itex]1^\infty[/itex] to be indeterminate. The reason we do this, is because of the limit

    [tex]\lim_{n\rightarrow +\infty} \left(1+\frac{1}{n}\right)^n[/tex]

    This will converge to "[itex]1^\infty[/itex]" in a certain sense, but it will not converge to 1. The reason that it does not converge to 1 is that [itex]1+\frac{1}{n}[/itex] converges to 1 too slow and that n converges to infinity too fast.

    Another explanation is the following, say that f(n) converges to 1 and that g(n) converges to infinity, then

    [tex]\lim_{n\rightarrow +\infty} f(n)^{g(n)} = \lim_{n\rightarrow +\infty} e^{g(n) log(f(n))}=e^{\lim_{n\rightarrow \infty} g(n) log(f(n))}[/tex]

    But as f(n) converges to 1, we have that log(f(n)) converges to 0. So [itex]g(n)log(f(n))[/itex] converges to "[itex]0\cdot (+\infty)[/itex]" and this is a known indeterminate form.

    So if we want to define [itex]1^\infty[/itex] and if we want to let it behave like we want it to, then we would have to define [itex]0\cdot (+\infty)[/itex] a well!!
    Last edited by a moderator: May 10, 2012
  4. May 10, 2012 #3
    Aha!! That clears it up.

    Thank you.
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