how would I show that $\displaystyle \lim_{x\to \infty}\frac{2^x}{3^{2x+1}}$ exists and what it is?
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.
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):
$\displaystyle n>N \implies a^n<a^N{\color{red}<}a^{\log_a\epsilon}=\epsilon$