<s>[INST] 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.
 [/INST] 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 $2/3$ 
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