Limit of this sequence $ \lim_{n\to \infty}\sqrt[n]{\frac{2^n + 3^n + 4^n}{5^n + 6^n}}$? How to find the limit of this sequence. I tried using the squeeze theorem but no result yet.
$ \lim_{n\to \infty}\sqrt[n]{\frac{2^n + 3^n + 4^n}{5^n + 6^n}}$
 A: It is easy to see that $$\frac{1}{2}(\frac{4^n}{6^n})<\frac{2^n+3^n+4^n}{5^n+6^n}<3(\frac{4^n}{6^n}).$$ Therefore,
$$\frac{4}{6}\sqrt[n]{\frac{1}{2}}<(\frac{2^n+3^n+4^n}{5^n+6^n})^\frac{1}{n}<\sqrt[n]{3}\frac{4}{6}.$$ Since $\sqrt[n]{c}\to 1$ as $n\to\infty$, we have that $$\lim_{n\to\infty}(\frac{2^n+3^n+4^n}{5^n+6^n})^\frac{1}{n}=\frac{4}{6}=\frac{2}{3}.$$
A: The limit is $\frac{4}{6} = \frac{2}{3}$. Prove that this is so.
HINT: Check that for each $\epsilon > 0$, there is an $n(\epsilon)$ such that for all $n>n(\epsilon)$, the denominator $2^n+3^n+4^n$ satisfies $$(1-\epsilon)4^n \le 2^n+3^n+4^n \le (1+\epsilon)4^n.$$ Likewise, there is an $n'(\epsilon)$ such that for all $n>n'(\epsilon)$, the numerator $5^n+6^n$ satisfies $$(1-\epsilon)6^n \le 5^n+6^n\le (1+\epsilon)6^n.$$
So for each $\epsilon > 0$ and for all $n > \max\{n(\epsilon), n'(\epsilon)\}$ the followign holds:
$$\sqrt[n]{\frac{1-\epsilon}{1+\epsilon}}\sqrt[n]{\frac{4^n}{6^n}} \le \sqrt[n]{\frac{5^n+6^n}{2^n+3^n+4^n}} \le \sqrt[n]{\frac{1+\epsilon}{1-\epsilon}}\sqrt[n]{\frac{4^n}{6^n}}$$
Can you finish from here.
