How can I evaluate: $\lim_{n\to \infty} \frac{\ln(n^2+n+1)}{\ln(n^3+n+1)}$? How should I go about finding $$\lim_{n\to \infty} \frac{\ln(n^2+n+1)}{\ln(n^3+n+1)}$$ Do I have to apply the log base change formula (although it didn't seem to resolve anything)? 
 A: $$
\begin{align}
\lim_{n\to\infty}\frac{\log(n^2+n+1)}{\log(n^3+n+1)}
&=\lim_{n\to\infty}\frac{2\log(n)+\log\left(1+\frac1n+\frac1{n^2}\right)}{3\log(n)+\log\left(1+\frac1{n^2}+\frac1{n^3}\right)}\\
&=\lim_{n\to\infty}\frac{2+\frac{\log\left(1+\frac1n+\frac1{n^2}\right)}{\log(n)}}{3+\frac{\log\left(1+\frac1{n^2}+\frac1{n^3}\right)}{\log(n)}}\\
&=\frac{2+\lim\limits_{n\to\infty}\frac{\log\left(1+\frac1n+\frac1{n^2}\right)}{\log(n)}}{3+\lim\limits_{n\to\infty}\frac{\log\left(1+\frac1{n^2}+\frac1{n^3}\right)}{\log(n)}}\\[12pt]
&=\frac23
\end{align}
$$
A: As $n\to\infty$, $\log(n^2 + n + 1) \sim 2\log(n)$ and 
$\log(n^3 + n + 1) \sim 3\log(n)$, so the limit of the quotient is $2/3$. 
A: Use  equivalents:
We have $n^2+n+1\sim_\infty n^2$, hence, as it does  not approach $1$, $\ln(n^2+n+1)\sim_\infty\ln(n^2)=2\ln n$.
Similarly, $\ln(n^3+n+1)\sim_\infty\ln(n^3)=3\ln n$. So
$$\frac{\ln(n^2+n+1)}{\ln(n^3+n+1)}\sim_\infty\frac{2\ln n}{3\ln n}=\frac23.$$
A: Every answer so far is amazing, I just wanna show how to do it with L'Hospital, (may be a little bit lengthy).
$$\lim_{n\to \infty} \frac{\ln(n^2+n+1)}{\ln(n^3+n+1)}=\lim_{n\to \infty}\frac{\frac{2n+1}{n^2+n+1}}{\frac{3n^2+1}{n^3+n+1}}=\lim_{n\to \infty}\frac{(2+\frac{1}{n})(1+\frac{1}{n^2}+\frac{1}{n^3})}{(3+\frac{1}{n^2})(1+\frac{1}{n}+\frac{1}{n^2})}=\frac{2}{3}$$
A: Since $n\to\infty$ . Therefore     $\ln(n^2+n+1)~2\ln(n)$ and $\ln(n^3+n+1)-3\ln(n)$. So easily we can say that the answer is 2/3
