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Limits of exponential functions

  1. Sep 22, 2008 #1

    can I do the following?

    If I am asked to find as lim x -> inf of (3^n/2^n), can I do:

    3^n = e^(n*ln3)
    2^n = e^(n*ln2)

    assume C = lim (3^n/2^n).

    C = lim e^(n*ln3)/e^(n*ln2)
    ln C = lim ln [(exp(n*ln3)/(exp(n*ln2)]
    In C = lim ln(exp(n*ln3)) - ln(exp(n*ln2))
    ln C = lim n*ln3 - n*ln2
    ln C = lim n*(ln3 - ln2)

    The right hand side is clearly inf, and hence ln C = inf, which would mean that C = inf?
    Though wouldn't C = 0 also give you infinity? So how would I know which one it is?

  2. jcsd
  3. Sep 22, 2008 #2


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    ln(0) isn't infinity. It's undefined. n*(ln3-ln2) is clearly unbounded. So C=exp(n*(ln3-ln2)) is unbounded.
  4. Sep 23, 2008 #3
  5. Sep 23, 2008 #4


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    Why do it that way?
    [tex]\frac{3^n}{2^n}= \left(\frac{3}{2}\right)^n[/tex]
    and that, since 3/2> 1, clearly increases without bound as x increases.

    On the other hand,
    [tex]\frac{2^n}{3^n}= \left(\frac{2}{3}\right)^n[/tex]
    since 2/3< 1, clearly goes to 0.

    Doing it the "hard way", if y= 2n/3n, then ln(y)= ln(2n)- ln(3n)= n(ln(2)- ln(3)). ln(2)- ln(3) is negative so, as n goes to infinity, n(ln(2)- ln(3)) goes to negative infinity and, as Dick said, ln(0)= negative infinity.
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