# Math Help - yet another limit proof

1. ## yet another limit proof

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

2. Originally Posted by binkypoo
how would I show that $\lim_{x\to \infty}\frac{2^x}{3^{2x+1}}$ exists and what it is?
The expression you're taking the limit of can be written $\frac{1}{3} \cdot \left(\frac{2}{3^2}\right)^x$ and the limiting value should be apparent. Now I suppose you might have to give an epsilon-delta proof of this limit.

3. more generally, how could I prove that $\lim_{x\to \infty} a^x = 0$ when $0\leq a \leq 1$?

4. Originally Posted by dannyboycurtis
more generally, how could I prove that $\lim_{x\to \infty} a^x = 0$ when $0\leq a \leq 1$?
Convergence and the $\epsilon-N$ criterion: $\lim_{n \rightarrow +\infty} S_n= L$ iff for all $\epsilon > 0$ there exists $N > 0$ such that $n > N$ implies $|S_n - L| < \epsilon$.

Are you familiar with this.

5. Originally Posted by mr fantastic
Convergence and the $\epsilon-N$ criterion: $\lim_{n \rightarrow +\infty} S_n= L$ iff for all $\epsilon > 0$ there exists $N > 0$ such that n>N implies $|S_n - L| < \epsilon$.

Are you familiar with this.
I added a missing part to the definition above.

6. the problem Im having is finding an appropriate N.

7. Originally Posted by dannyboycurtis
the problem Im having is finding an appropriate N.
For $0, let $N=\log_a\epsilon$. Therefore,

$n>N \implies a^n

8. Originally Posted by redsoxfan325
For $0, let $N=\log_a\epsilon$. Therefore,

$n>N \implies a^n
Actually, one small adjustment: since $\log_a\epsilon$ is unlikely to be a natural number, you need to take $N=\lceil\log_a\epsilon\rceil$, where $\lceil x\rceil$ is the smallest integer greater than $x$. Then you have basically the same thing with one tiny adjustment (in red):

$n>N \implies a^n