Very simple lim

1. Oct 26, 2013

phospho

find the limit as n goes to infinity by using the sandwich theorem

$(3^n + 5^n + 7^n )^{\frac{1}{n}}$

I notice the limit is 7 by using a different method (not the sandwich theorem)

using the sandwich theorem I see that

$7^n \leq 3^n + 5^n + 7^n$ but I can't seem to find a good upper bound. I can see that $2.7^n$ works but I can't explain why $7^n \leq 3^n + 5^n + 7^n \leq 2.7^n$

2. Oct 26, 2013

LCKurtz

Try using $7^n\le 3^n + 5^n + 7^n \le 7^n+7^n+7^n$.

3. Oct 26, 2013

Staff: Mentor

I haven't worked the problem, but maybe this will help.
$3 \cdot 3^n \leq 3^n + 5^n + 7^n \leq 3 \cdot 7^n$

$\Rightarrow \lim_{n \to \infty} (3 \cdot 3^n)^{1/n} \leq \lim_{n \to \infty}(3^n + 5^n + 7^n)^{1/n} \leq \lim_{n \to \infty} (3 \cdot 7^n)^{1/n}$

4. Oct 26, 2013

phospho

so simple, thanks

I have another limits question (I've done the problem but would like someone to check it over as I'm new to limits). Should I start a new thread, or post it here? And is there a maximum amount of threads allowed per day?

5. Oct 26, 2013