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I know that 1^(+/- infinity) is indeterminate.Is c^(+/- infinity)

  1. Jul 7, 2010 #1
    I know that 1^(+/- infinity) is indeterminate.
    Is c^(+/- infinity) indeterminate for real numbers c?
     
  2. jcsd
  3. Jul 7, 2010 #2

    jbunniii

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    Re: c^(infinity)

    Why do you say that [itex]1^\infty[/itex] and [itex]1^{-\infty}[/itex] are indeterminate? We have

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

    and

    [tex]\lim_{n \rightarrow -\infty} 1^n = 1[/tex]

    If [itex]c > 1[/itex], then

    [tex]\lim_{n \rightarrow \infty} c^n = \infty[/tex]

    and

    [tex]\lim_{n \rightarrow -\infty} c^n = 0[/tex]

    and the opposite result holds for [itex]0 < c < 1[/itex]

    One of the limits doesn't exist if [itex]c < -1[/itex], and the other one doesn't exist if [itex]-1 < c < 0[/itex]. Neither limit exists if [itex]c = -1[/itex]. (I'll let you work out the details.)
     
  4. Jul 7, 2010 #3

    Mark44

    Staff: Mentor

    Re: c^(infinity)

    Think about this in terms of limits. Do you have a feel for what the values of these limits are?
    [tex]\lim_{n \to \infty} 2^n[/tex]
    [tex]\lim_{n \to \infty} 2^{-n} = \lim_{n \to \infty} \frac{1}{2^n}[/tex]

    To investigate this further for arbitrary bases, you probably want to limit the base to the positive reals. One case would be for a > 1. Another would be for 0 < a < 1.
     
  5. Jul 7, 2010 #4

    Mark44

    Staff: Mentor

    Re: c^(infinity)

    One of the indeterminate forms is [itex][1^{\infty}][/itex]. An example of this is [tex]\lim_{x \to 0} (1 + x)^{1/x}[/tex].

    Here we have the base approaching 1 and the exponent growing large without bound, yet the limit is not 1.
     
  6. Jul 7, 2010 #5
    Re: c^(infinity)

    Heh. Gotcha, thanks.
     
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