# Math Help - limit with exponent variable

1. ## 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.

2. $\frac{2^n}{3^{n+1}} = \frac{2^n}{3 \cdot 3^n} = \frac{1}{3} \cdot (\frac{2}{3})^n$

3. and by definition $(\frac{2}{3})^n$ goes to 0 because multiplying a fraction by itself many times makes a smaller number?

4. Yes, this is the point, but it's true because $\frac{2}{3} < 1$, this is important.

5. 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!