<s>[INST] 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?
 [/INST] 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).
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