Compute the limit of $n\cdot\left(\frac{\arccos\left(3/n^3\right)}{\arccos\left(3/(n+1)^3\right)}-1\right)$ when $n\to\infty$ How to solve limit like this? 

$$\lim_{n\to\infty}{n\left(\frac{\arccos\frac3{n^3}}{\arccos\frac3{(n+1)^3}}-1\right)}$$

$$=\lim_{n\to\infty}{n\left(\frac{\frac\pi2}{\frac\pi2}-1\right)}$$

$$=\lim_{n\to\infty}{n(1-1)} = 0$$
but $\infty*0$ is not define, how to solve it?
This is just a naive thought, but have you tried the first order approximations $\arccos\left(\frac{3}{n^3}\right)\approx\frac{\pi}{2}-\frac{3}{n^3}$ and $\arccos\left(\frac{3}{(n+1)^3}\right)\approx\frac{\pi}{2}-\frac{3}{(n+1)^3}$ for $n\rightarrow\infty$? In this way the whole expression becomes
$$-\frac{18n^2+18n+6}{\pi n^5+3\pi n^4+3\pi n^3+(\pi+3)n^2},$$
whose limit is $0$ (as it should be).
Suppose $A\to \infty$ and $B\to 0$ as $n\to \infty$. If you want to solve $\lim_{n\to\infty} AB$, you can rewrite it as $\lim_{n\to\infty} \frac{A}{1/B}$ to get the indeterminate form $\frac{\infty}{\infty}$ in the limit, or as $\lim_{n\to\infty}\frac{B}{1/A}$ to get the indeterminate form $\frac{0}{0}$ in the limit. Either of these can then be solved using L'hopital's rule.