Weyl's law states the eigenvalues of the Laplacian behave as $$\lambda_j \sim f(n)j^{\frac 2n}\quad\text{as $j \to \infty$}$$ where $n$ is the dimension.
Does this literally mean that, $$\lim_{j \to \infty}\frac{\lambda_j}{f(n)j^{\frac 2n}} = 1?$$
Because I want to conclude that there is a constant $A=A(n)$ such that $\frac{\lambda_{j+1}}{\lambda_j} \leq A$ for all $j$. Am I right that this holds, since that is what the asymptotic formula means?
Yes. Note that $f$ is a function not just of dimension but of the volume of the domain.