How to solve $\lim\limits_{n \to +\infty} \left(\frac{n \ln(n^3+5n)}{\ln(n^{3n} +n^7)}\right)$?

Question

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

$$\lim\limits_{n \to +\infty} \left(\frac{n \ln(n^3+5n)}{\ln(n^{3n} +n^7)}\right)$$

Answer 1

$$\lim_{n \to +\infty} \frac{n\ln(n^3+5n)}{\ln(n^{3n} +n^7)} =\lim_{n \to +\infty} \frac{n\ln\left(n^3\left(1+\frac{5n}{n^3}\right)\right)}{\ln\left(n^{3n}\left(1+\frac{n^7}{n^{3n}}\right)\right)} =\lim_{n \to +\infty} \frac{n\ln(n^3)}{\ln(n^{3n})} =\lim_{n \to +\infty} \frac{3n\ln(n)}{3n\ln(n)}=1$$

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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}

Answer 2

With equivalents, it is very short: $$\frac{n\ln(n^3+5n)}{\ln(n^{3n}+n^7)}\sim_\infty\frac{n\ln n^3}{\ln n^{3n}}=\frac{n\cdot3\ln n}{3n\ln n}=1.$$