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).
We can proceed as follows \begin{align} L &= \lim_{n \to \infty}n\left(\dfrac{\arccos\left(\dfrac{3}{n^{3}}\right)}{\arccos\left(\dfrac{3}{(n + 1)^{3}}\right)} - 1\right)\notag\\ &= \lim_{n \to \infty}n\left(\dfrac{\arcsin\left(\dfrac{3}{(n + 1)^{3}}\right) - \arcsin\left(\dfrac{3}{n^{3}}\right)}{\dfrac{\pi}{2} - \arcsin\left(\dfrac{3}{(n + 1)^{3}}\right)}\right)\notag\\ &= \frac{2}{\pi}\lim_{n \to \infty}n\left(\arcsin\left(\dfrac{3}{(n + 1)^{3}}\right) - \arcsin\left(\dfrac{3}{n^{3}}\right)\right)\notag\\ &= \frac{2}{\pi}\lim_{n \to \infty}n\arcsin\left(\dfrac{3}{(n + 1)^{3}}\cdot\sqrt{1 - \dfrac{9}{n^{6}}} - \frac{3}{n^{3}}\sqrt{1 - \frac{9}{(n + 1)^{6}}}\right)\notag\\ &= \frac{2}{\pi}\lim_{n \to \infty}n\arcsin f(n)\notag \end{align} The function $f(n)$ is algebraic and it is easily proved that $n f(n) \to 0$ and hence $n\arcsin f(n) = nf(n)((\arcsin f(n))/f(n)) \to 0 \cdot 1 = 0$ so the desired limit is $0$.
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.
