how would I show that $\displaystyle \lim_{x\to \infty}\frac{2^x}{3^{2x+1}}$ exists and what it is?

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- Oct 17th 2009, 08:04 PMbinkypooyet another limit proof
how would I show that $\displaystyle \lim_{x\to \infty}\frac{2^x}{3^{2x+1}}$ exists and what it is?

- Oct 17th 2009, 08:21 PMmr fantastic
- Oct 19th 2009, 07:13 PMdannyboycurtis
more generally, how could I prove that $\displaystyle \lim_{x\to \infty} a^x = 0$ when $\displaystyle 0\leq a \leq 1$?

- Oct 19th 2009, 07:24 PMmr fantastic
**Convergence and the $\displaystyle \epsilon-N$ criterion:**$\displaystyle \lim_{n \rightarrow +\infty} S_n= L$ iff for all $\displaystyle \epsilon > 0$ there exists $\displaystyle N > 0$ such that $\displaystyle n > N$ implies $\displaystyle |S_n - L| < \epsilon$.

Are you familiar with this. - Oct 19th 2009, 07:30 PMredsoxfan325
- Oct 19th 2009, 07:39 PMdannyboycurtis
the problem Im having is finding an appropriate N.

- Oct 19th 2009, 07:50 PMredsoxfan325
- Oct 19th 2009, 08:33 PMredsoxfan325
Actually, one small adjustment: since $\displaystyle \log_a\epsilon$ is unlikely to be a natural number, you need to take $\displaystyle N=\lceil\log_a\epsilon\rceil$, where $\displaystyle \lceil x\rceil$ is the smallest integer greater than $\displaystyle x$. Then you have basically the same thing with one tiny adjustment (in red):

$\displaystyle n>N \implies a^n<a^N{\color{red}<}a^{\log_a\epsilon}=\epsilon$