Neither of those. For $n>1$, $\lfloor\frac{1}{n}\rfloor=0$. Hence, $\lfloor\frac{1}{n}\rfloor=0=O(0)$.

Also, you can easily rule out $\Omega(\log(n))$ since $\log$ is an increasing function, while $\frac{1}{n}$ is a decreasing function.

Neither of those. For $n>1$, $\lfloor\frac{1}{n}\rfloor=0$. Hence, $\lfloor\frac{1}{n}\rfloor=0=O(0)$.

Also, you can easily rule out $\Omega(\log(n))$ since $\log$ is an increasing to infinity, while $\frac{1}{n}$ is a decreasing function.