# yet another limit proof

• Oct 17th 2009, 08:04 PM
binkypoo
yet 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 PM
mr fantastic
Quote:

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

The expression you're taking the limit of can be written $\displaystyle \frac{1}{3} \cdot \left(\frac{2}{3^2}\right)^x$ 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, 07:13 PM
dannyboycurtis
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 PM
mr fantastic
Quote:

Originally Posted by dannyboycurtis
more generally, how could I prove that $\displaystyle \lim_{x\to \infty} a^x = 0$ when $\displaystyle 0\leq a \leq 1$?

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 PM
redsoxfan325
Quote:

Originally Posted by mr 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 n>N implies $\displaystyle |S_n - L| < \epsilon$.

Are you familiar with this.

I added a missing part to the definition above.
• Oct 19th 2009, 07:39 PM
dannyboycurtis
the problem Im having is finding an appropriate N.
• Oct 19th 2009, 07:50 PM
redsoxfan325
Quote:

Originally Posted by dannyboycurtis
the problem Im having is finding an appropriate N.

For $\displaystyle 0<a<1$, let $\displaystyle N=\log_a\epsilon$. Therefore,

$\displaystyle n>N \implies a^n<a^N=a^{\log_a\epsilon}=\epsilon$
• Oct 19th 2009, 08:33 PM
redsoxfan325
Quote:

Originally Posted by redsoxfan325
For $\displaystyle 0<a<1$, let $\displaystyle N=\log_a\epsilon$. Therefore,

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

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):

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