<s>[INST] $\lim\limits_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))}$ I want to solve this limit:
$$\lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))}$$
I have proved that $\lim\limits_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n} = 0$ and $\lim\limits_{n \rightarrow +\infty} \frac{1}{(1-\cos(1/n^2))}= \infty$ but I have indeterminate form. How can I solve that?
 [/INST] A long version:
$$\lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))}=
\lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-\ln(n^3)}{n\cdot\frac{1}{n^4}\cdot\frac{(1-\cos^2(1/n^2))}{\frac{1}{n^4}}}\cdot(1+\cos(1/n^2))=\\
\lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-\ln(n^3)}{\frac{1}{n^3}\cdot \color{red}{\frac{\sin^2(1/n^2)}{\frac{1}{n^4}}}}\cdot(1+\cos(1/n^2))=\\
2\cdot \lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-\ln(n^3)}{\frac{1}{n^3}}=
2\cdot \lim_{n \rightarrow +\infty} n^3\ln\left(\frac{1+n+n^3}{n^3}\right)=\\
2\cdot \lim_{n \rightarrow +\infty} \ln\left(1+\frac{1+n}{n^3}\right)^{n^3}=\\
2\cdot \lim_{n \rightarrow +\infty} \ln\left(1+\frac{1+n}{n^3}\right)^{\frac{n^3}{n+1}\cdot (n+1)} = \\
2\cdot \lim_{n \rightarrow +\infty} (n+1)\cdot\ln\left(1+\frac{1+n}{n^3}\right)^{\frac{n^3}{n+1}} =
2\cdot \ln{e} \cdot \lim_{n \rightarrow +\infty} (n+1)\rightarrow +\infty$$

On the 2nd line
$$\lim_{n \rightarrow +\infty}\frac{\sin^2(1/n^2)}{\frac{1}{n^4}}=
\lim_{n \rightarrow +\infty}\frac{\sin(1/n^2)}{\frac{1}{n^2}}\cdot \frac{\sin(1/n^2)}{\frac{1}{n^2}}=1$$ from
$$\lim\limits_{x\rightarrow0}\frac{\sin x}{x}=1$$
</s>