How to solve $\lim\limits_{n \to +\infty} \left(\frac{n \ln(n^3+5n)}{\ln(n^{3n} +n^7)}\right)$? I have a problem with this limit, I don't know what method to use.
Can you show a method for the resolution with asymptotic approximations(so without Hopital)? Thanks
$$\lim\limits_{n \to +\infty} \left(\frac{n \ln(n^3+5n)}{\ln(n^{3n} +n^7)}\right)$$
 A: $$\lim_{n \to +\infty} \frac{n\ln(n^3+5n)}{\ln(n^{3n} +n^7)}
=\lim_{n \to +\infty} \frac{n\ln\left(n^3\left(1+\frac{5n}{n^3}\right)\right)}{\ln\left(n^{3n}\left(1+\frac{n^7}{n^{3n}}\right)\right)}
=\lim_{n \to +\infty} \frac{n\ln(n^3)}{\ln(n^{3n})}
=\lim_{n \to +\infty} \frac{3n\ln(n)}{3n\ln(n)}=1$$

More rigorous version for the bracket:
\begin{align}
\lim_{n \to +\infty} \frac{n\ln(n^3+5n)}{\ln(n^{3n} +n^7)}
&=\lim_{n \to +\infty} \frac{n\ln\left(n^3\left(1+\frac{5n}{n^3}\right)\right)}{\ln\left(n^{3n}\left(1+\frac{n^7}{n^{3n}}\right)\right)}\\
&=\lim_{n \to +\infty} \frac{n\ln(n^3)+n\ln\left(1+\frac{5n}{n^3}\right)}{\ln(n^{3n})+\ln\left(1+\frac{n^7}{n^{3n}}\right)}\\
&=\lim_{n \to +\infty} \frac{n\ln(n^3)+\frac5n\cdot\frac{n^3}{5n}\ln\left(1+\frac{5n}{n^3}\right)}{\ln(n^{3n})+\ln\left(1+\frac{n^7}{n^{3n}}\right)}\\
&=\lim_{n \to +\infty} \frac{n\ln(n^3)+\frac5n}{\ln(n^{3n})}\\
&=\lim_{n \to +\infty} \frac{n\ln(n^3)}{\ln(n^{3n})}\\
&=\lim_{n \to +\infty} \frac{3n\ln(n)}{3n\ln(n)}\\&=1
\end{align}
A: With equivalents, it is very short:
$$\frac{n\ln(n^3+5n)}{\ln(n^{3n}+n^7)}\sim_\infty\frac{n\ln n^3}{\ln n^{3n}}=\frac{n\cdot3\ln n}{3n\ln n}=1.$$
