Limit of $(\frac{2^n+3^n+4^n}{5^n+6^n})^{\frac{1}{n}}$ I assume it is $1$. 
Though $1$ is a limit of the function  $(\frac{2^x+3^x+4^x}{5^x+6^x})^{\frac{1}{x}}$. Besides, even if it is $1$, I still need to solve the following inequality
$|(\frac{2^n+3^n+4^n}{5^n+6^n})^{\frac{1}{n}}-1|< \epsilon$ 
for arbitrary $\epsilon > 0$. That I don't know how to do as well.
 A: Notice that 
$$(\frac{2^n+3^n+4^n}{5^n+6^n})^{\frac{1}{n}} \geq (\frac{4^n}{2 \cdot 6^n})^{\frac{1}{n}} = \frac{2}{3}\cdot (\frac{1}{2})^{\frac{1}{n}}$$
While
$$(\frac{2^n+3^n+4^n}{5^n+6^n})^{\frac{1}{n}} \leq (\frac{3 \cdot 4^n}{6^n})^{\frac{1}{n}}=\frac{2}{3}\cdot 3^{\frac{1}{n}}$$
And
$$\lim \limits_{n \to \infty} (\frac{1}{2})^{\frac{1}{n}}=\lim \limits_{n \to \infty} 3^{\frac{1}{n}}=1.$$
So the answer is $\frac{2}{3}$.
However I assume you are a beginner like me, thus maybe prefer to prove it by definition. It is not difficult, you only need to repeat trick I have shown.
$$|(\frac{2^n+3^n+4^n}{5^n+6^n})^{\frac{1}{n}}-\frac{2}{3}| \leq |(\frac{3 \cdot 4^n}{6^n})^{\frac{1}{n}}-\frac{2}{3}| \\
=\frac{2}{3} \cdot (3^{\frac{1}{n}}-1)$$
Thus for any positive real number $\epsilon$, there is a positive integer $N=\max \{1, \lfloor \frac{\ln 3}{\ln (\frac{3\epsilon}{2}+1)} \rfloor \}$ such that when $n>N$, $$|(\frac{2^n+3^n+4^n}{5^n+6^n})^{\frac{1}{n}}-\frac{2}{3}|<\epsilon.$$
That's what we want to prove.
Edited: So far I think the most nature way in this kind of approach is to notice the result here: Prove the following limit $\lim \limits_{n \to \infty} (3^n+4^n)^{\frac{1}{n}} =4$.
I think one can prove it without squeeze lemma.
So
$$\lim \limits_{n \to \infty} (2^n+3^n+4^n)^{\frac{1}{n}}=4$$
And
$$\lim \limits_{n \to \infty} (5^n+6^n)^{\frac{1}{n}}=6$$
Thus the original limit is $\frac{2}{3}$.
A: Here is a more classical approach. Take the $ln$ on both sides to get: $\frac{1}{n}*ln\frac{2^n+3^n+4^n}{5^n+6^n}$=$\frac{ln(2^n+3^n+4^n)-ln(5^n+6^n)}{n}$=$\frac{ln(2^n+3^n+4^n)}{n}-\frac{ln(5^n+6^n)}{n}$ With $n$ to infinity, we can apply L'Hospital's Rule: $\frac{2^nln2+3^nln3+4^nln4}{2^n+3^n+4^n}-\frac{5^nln5+6^nln6}{5^n+6^n}$ Dividing all terms in the first fraction by $4^n$ and the second fraction by $6^n$ and let $n$ go to infinity, we arrive at $ln4-ln6$ which is $ln\frac{2}{3}$ (please verify!) Since this is really the $ln$ of our limit, the actual limit is $2/3$ 
