# yet another limit proof

• Oct 17th 2009, 09:04 PM
binkypoo
yet another limit proof
how would I show that $\lim_{x\to \infty}\frac{2^x}{3^{2x+1}}$ exists and what it is?
• Oct 17th 2009, 09:21 PM
mr fantastic
Quote:

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.
• Oct 19th 2009, 08:13 PM
dannyboycurtis
more generally, how could I prove that $\lim_{x\to \infty} a^x = 0$ when $0\leq a \leq 1$?
• Oct 19th 2009, 08:24 PM
mr fantastic
Quote:

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.
• Oct 19th 2009, 08:30 PM
redsoxfan325
Quote:

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.
• Oct 19th 2009, 08:39 PM
dannyboycurtis
the problem Im having is finding an appropriate N.
• Oct 19th 2009, 08:50 PM
redsoxfan325
Quote:

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
• Oct 19th 2009, 09:33 PM
redsoxfan325
Quote:

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