$\lim\limits_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))}$ 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?
Let's write
$$\frac{\ln(1+n+n^3) - 3 \ln(n)}{n(1- \cos(1/n^2))} = \frac{\ln(1 + \frac{1}{n^2} + \frac{1}{n^3})}{n(1- \cos(1/n^2))} = \frac{\frac{1}{n^2} + \frac{1}{n^3} +o(\frac{1}{n^3})}{n (\frac{1}{2n^4} + o(\frac{1}{n^5}))} = \frac{n +1 + o(1)}{\frac{1}{2} + o(\frac{1}{n})} \sim 2n$$

so the limit is $+\infty$.
I would be inclined to write this as $\lim_{x\to\infty}\frac{\ln(1+ x+ x^3)- 3 \ln(x)}{x(1+ \cos(1/x^2)}$ where $x$ does not have to be an integer. That way the functions are differentiable and I can use "L'Hopital's rule".

We can write the numerator as $\ln(1+ x + x^3)- 3 \ln(x)= \ln(1+ x+ x^3)- \ln(x^3)= \ln\left(\frac{1 + x+ x^3}{x^3}\right)= \ln(x^{-3} + x^{-2} + 1)$.

Its derivative is $\frac{-3x^{-4}- 2x^{-3}}{x^{-3}+ x^{-2}+ 1}= \frac{-3- 2x}{x+ x^2+ x^3}$. Since the denominator has higher degree than the numerator, the limit as $n$ (and so $x$) goes to infinity is $0$.

We can write the denominator as $x(1+ \cos(1/x))= x(1+ \cos(x^{-1}))$.

Its derivative is $1+ \cos(x^{-1})+ x(-\sin(x^{-1}))(-x^{-2})= 1+ \cos(x^{-1})- x^{-1}\sin(x^{-1})$.

As $x$ goes to infinity all those "$x^{-1}$" terms go to $0$ so the denominator goes to $1+ 1- 0= 2$.

Since that denominator is nonzero, the limit of the original expression is $0$.