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Indeterminate Forms

  1. Aug 11, 2006 #1
    Why is [tex]1^\infty[/tex] an indeterminate form? If you keep multiplying 1 by itself doesn't the answer stay 1?
  2. jcsd
  3. Aug 11, 2006 #2
    It is an indeterminate form because [tex]0 . \infty[/tex] is.

    Try to take the "logarithm of [tex]1^\infty[/tex]".

    Obviously it is not a proof.
    It is only a mnemonic trick.
    Last edited: Aug 11, 2006
  4. Aug 11, 2006 #3


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    [tex]\lim_{x\rightarrow \infty} 1^x=1[/tex].

    But consider a case like

    [tex]\lim_{x\rightarrow \infty} (1+\frac{1}{x})^{e^{x^5}}[/tex]

    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" [itex]1^{\infty}[/tex] will diverge, and some other times, it will behave and go to 1.

    Maybe in some cases, the limit will converge to a value that is not 1? Can someone provide an exemple?
    Last edited: Aug 11, 2006
  5. Aug 11, 2006 #4
    And, obviously,

    [tex]\lim_{x\rightarrow \infty} 0 x = 0[/tex]

    [tex]\lim_{x\rightarrow \infty} (\cos(1/x^2))^{x^4}[/tex]
    Last edited: Aug 11, 2006
  6. Aug 11, 2006 #5
    Yeah, that's what I was talking about. I know that the form [tex]1^{\infty}[/tex] has many different values; I wanted to know why.
    Can you run through the "converging at different speeds" bit a lot slower (pun unintended, but noted and pride taken)?
  7. Aug 11, 2006 #6


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    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.
    Last edited: Aug 11, 2006
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