imit \( \lim _{n \rightarrow \infty} \frac{\sqrt{4 n^{3}-n}}{2 n \sqrt{n-1}} \)

First, let's simplify the expression inside the limit: \[ \lim_{n \rightarrow \infty} \frac{\sqrt{4n^3 - n}}{2n \sqrt{n-1}}. \] As \( n \) becomes very large, the term \( 4n^3 \) will dominate \( -n \) in the square root. Thus, we can approximate: \[ \sqrt{4n^3 - n} \approx \sqrt{4n^3} = 2n^{3/2}. \] Now, simplifying the denominator: \[ \sqrt{n - 1} \approx \sqrt{n} \quad \text{as } n \to \infty, \] so, \[ 2n \sqrt{n - 1} \approx 2n \sqrt{n} = 2n^{3/2}. \] Now, substituting these approximations back into the limit yields: \[ \lim_{n \rightarrow \infty} \frac{2n^{3/2}}{2n^{3/2}} = \lim_{n \rightarrow \infty} 1 = 1. \] Therefore, the limit evaluates to: \[ \boxed{1}. \]