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Infinite Limit

  1. Feb 25, 2006 #1
    How do I evaluate this limit?

    \mathop {\lim }\limits_{n \to \infty } \left( {\frac{2}{3}} \right)^n

    Is this the correct approach?

    {\rm{Let}} \; \; \; y = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{2}{3}} \right)^n

    \ln y = \mathop {\lim }\limits_{n \to \infty } \ln \left[ {\left( {\frac{2}{3}} \right)^n } \right]

    \ln y = \mathop {\lim }\limits_{n \to \infty } \left[ {n \cdot \ln \left( {\frac{2}{3}} \right)} \right]

    I am stuck at this step. I don't see a way to manipulate the limit into a
    form that L'Hopital's Rule will apply. I know the limit evaluates to 0.
    Last edited: Feb 25, 2006
  2. jcsd
  3. Feb 25, 2006 #2


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    Homework Helper

    No l'Hospital's rule needed. Since [tex]\frac{2}{3} < 1,\left( \frac{2}{3}\right) ^{n}\rightarrow 0 \mbox{ as }n\rightarrow\infty[/tex]
  4. Feb 25, 2006 #3


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    Homework Helper

    You can continue by noticing that:
    [tex]\ln \left( \frac{2}{3} \right) < 0[/tex]
    So as [tex]n \rightarrow +\infty[/tex], [tex]n \star \ln \left( \frac{2}{3} \right) \rightarrow - \infty[/tex], right?
    So as [tex]n \rightarrow +\infty[/tex], [tex]\ln y \rightarrow - \infty[/tex]
    So what's [tex]y \rightarrow ?[/tex]
    Or as benorin has pointed out:
    If |a| < 1 then [tex]\lim_{n \rightarrow \infty} a ^ n = 0[/tex]
    If a = 1 then [tex]\lim_{n \rightarrow \infty} a ^ n = 1[/tex]
    Can you get this? :)
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