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Limit of a^x

  1. Oct 16, 2006 #1

    I was doing some exercises on limits when I stumbled upon the following questions:
    (limits are always x->infinity)

    lim (2^x+1)/(3^x+1)

    2nd one:
    lim (-9/8)^x

    The first one should be zero (3^x expands faster than 2^x), and the 2nd one doesn't seem to exist (according to maple and the solutions :wink: , I suppose it has something to do with the graph being all chopped up and not knowing if x->infinity is even or odd?).
    Is there a way to find these answers trough calculation instead of uncertain logic :tongue: ?
    Another exercise similar to the first one:
    lim (8/9)^x
  2. jcsd
  3. Oct 16, 2006 #2


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    [tex]a^x = \exp(\ln a^x) = \exp(x \ln a) = (e^x)^{\ln a}[/tex]

    you go from there.
  4. Oct 16, 2006 #3
    So lim(8/9)^x = lim e^(x*ln(8/9))
    ln(8/9) (or ln (8 ) -ln(9)) is smaller than 0, resulting in a negative value for x*(ln(8/9))
    So the result would be something of this form:
    lim e^(-x*...)
    lim 1/e^(x*...)
    going towards 0
    Am I correct?

    Now, the limit of (-9/8)^x can't be calculated this way unless we use irrational numbers ( ln(a) = ln(-a)+Pi*I for a<0 ) , right? But is this the proof that the limit doesn't exist? or is there another reason?
  5. Oct 16, 2006 #4


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    My hint was aimed at solving the first limit.

    For the second one, you could show that the sequence {(-9/8)^2n} goes to +infinity and the sequence {(-9/8)^(2n+1)} goes to -infinity.
  6. Oct 16, 2006 #5
    for the second one you could also show that it is a geometric sequence with [tex] |r| > 1 [/tex].
  7. Oct 17, 2006 #6


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    How, exactly, are you defining ax for a< 0?
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