$\lim\limits_{x \rightarrow \infty} \frac{\ln [x]}{x}=$

1 0 -1 does not exist

0

x - 1 ≤ [x]  ≤ x $\Rightarrow \frac{\ln (x-1)}{x} \leq \frac{\ln [x]}{x} \leq \frac{\ln x \mid}{x} \Rightarrow \lim\limits_{x \rightarrow \infty} \frac{\ln [x]}{x}=0$ Hence (2) is the correct answer.