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Math Help - limit with exponent variable

  1. #1
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    limit with exponent variable

    find the limit of a_n = \frac{2^n}{3^{n+1}} as n approaches infinity

    L'Hopitals rule wont work here. I do not know what technique to use to find the limit. I can look at the graph and tell that it is converging at 0.
    Last edited by diroga; June 9th 2009 at 01:44 AM.
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  2. #2
    Member Ruun's Avatar
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    \frac{2^n}{3^{n+1}} = \frac{2^n}{3 \cdot 3^n} = \frac{1}{3} \cdot (\frac{2}{3})^n

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  3. #3
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    and by definition (\frac{2}{3})^n goes to 0 because multiplying a fraction by itself many times makes a smaller number?
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  4. #4
    Member Ruun's Avatar
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    Yes, this is the point, but it's true because \frac{2}{3} < 1, this is important.
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  5. #5
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    Quote Originally Posted by diroga View Post
    and by definition (\frac{2}{3})^n goes to 0 because multiplying a fraction by itself many times makes a smaller number?
    Not "by definition" and saying "less than 1" as Ruun did is better than saying "a fraction". After all, \frac{3}{2} is a fraction!
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