# limit with exponent variable

• Jun 9th 2009, 02:31 AM
diroga
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.
• Jun 9th 2009, 02:37 AM
Ruun
$\frac{2^n}{3^{n+1}} = \frac{2^n}{3 \cdot 3^n} = \frac{1}{3} \cdot (\frac{2}{3})^n$

(Hi)
• Jun 9th 2009, 02:54 AM
diroga
and by definition $(\frac{2}{3})^n$ goes to 0 because multiplying a fraction by itself many times makes a smaller number?
• Jun 9th 2009, 03:03 AM
Ruun
Yes, this is the point, but it's true because $\frac{2}{3} < 1$, this is important.
• Jun 9th 2009, 05:36 AM
HallsofIvy
Quote:

Originally Posted by diroga
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!