$ \lim_{n \to \infty} \frac{\log n^3}{\log (n^3+3n^2)}$

Question

I want to solve this $$ \lim_{n \to \infty} \frac{\log n^3}{\log \left(n^3+3n^2\right)}$$ I found that $$ \frac{\log n^3}{\log \left(n^3+3n^2\right)}<\frac{\log\left( n^3\right)}{\log \left(n^3\right)}=1$$ but I need another bond according to the squeeze theorem.

Answer 1

$$=\lim_{n\to\infty}\dfrac{3\ln n}{3\ln n+\ln\left(1+\dfrac3n\right)}=\lim_{n\to\infty}\dfrac{3}{3+\dfrac{\ln\left(1+\dfrac3n\right)}{\ln n}}=?$$

Answer 2

If you use equivalents, it's obvious: $\;n^3+3n^2\sim_\infty n^3$, so $$\log(n^3+3n^2)\sim_\infty\log n^3,\quad\text{and finally }\;\frac{\log n^3}{\log(n^3+3n^2)}\sim_\infty\frac{\log n^3}{\log n^3}=1.$$