$\lim\limits_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))}$ 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?

You may proceed as follows:


*

*$\frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))} = \frac{\ln \left(1+\frac{1}{n^2}+\frac{1}{n^3} \right)}{n(1-\cos(1/n^2))}$
Now, set 


*

*$n = \frac{1}{x}$ and consider the limit for $x \to 0^+$ and use

*$\cos t > 1- \frac{t^2}{2}$ for $t \in (0,\frac{\pi}{2})$
So, you get
\begin{eqnarray*} \frac{\ln \left(1+\frac{1}{n^2}+\frac{1}{n^3} \right)}{n(1-\cos(1/n^2))}
& \stackrel{n = \frac{1}{x}}{=} & \frac{x\cdot \ln (1+x^2+x^3)}{1-\cos x^2}\\
& \stackrel{\cos x^2 > 1 - \frac{x^4}{2}}{>} & \frac{2\ln (1+x^2+x^3)}{x^3} \\
&  \stackrel{L'Hosp.}{\sim} & \frac{2}{1+x^2+x^3}\left(\frac{2}{x^2} + \frac{3}{x} \right) \\
&  \stackrel{x \to 0^+}{\longrightarrow} & +\infty
\end{eqnarray*}
