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

Evaluate the limit by following steps: - step0: Evaluate using transformations: \(\lim _{n\rightarrow +\infty}\left(\frac{\sqrt{4n^{3}-n}}{2n\sqrt{n-1}}\right)\) - step1: Rewrite the expression: \(\lim _{n\rightarrow +\infty}\left(\frac{\sqrt{4-\frac{1}{n^{2}}}\times \left|n\right|\times \sqrt{n}}{2\sqrt{1-\frac{1}{n}}\times \left|n\right|\times \sqrt{n}}\right)\) - step2: Reduce the fraction: \(\lim _{n\rightarrow +\infty}\left(\frac{\sqrt{4-\frac{1}{n^{2}}}}{2\sqrt{1-\frac{1}{n}}}\right)\) - step3: Rewrite the expression: \(\frac{\lim _{n\rightarrow +\infty}\left(\sqrt{4-\frac{1}{n^{2}}}\right)}{\lim _{n\rightarrow +\infty}\left(2\sqrt{1-\frac{1}{n}}\right)}\) - step4: Calculate: \(\frac{2}{\lim _{n\rightarrow +\infty}\left(2\sqrt{1-\frac{1}{n}}\right)}\) - step5: Calculate: \(\frac{2}{2}\) - step6: Reduce the numbers: \(\frac{1}{1}\) - step7: Calculate: \(1\) To evaluate the limit \[ \lim _{n \rightarrow \infty} \frac{\sqrt{4 n^{3}-n}}{2 n \sqrt{n-1}}, \] we can simplify the expression step by step. 1. **Analyze the numerator**: The term inside the square root in the numerator is \(4n^3 - n\). As \(n\) approaches infinity, the \(4n^3\) term dominates, so we can approximate: \[ \sqrt{4n^3 - n} \approx \sqrt{4n^3} = 2n^{3/2}. \] 2. **Analyze the denominator**: The denominator is \(2n\sqrt{n-1}\). As \(n\) approaches infinity, we can approximate \(\sqrt{n-1}\) as \(\sqrt{n}\): \[ \sqrt{n-1} \approx \sqrt{n}. \] Therefore, the denominator can be approximated as: \[ 2n\sqrt{n-1} \approx 2n\sqrt{n} = 2n^{3/2}. \] 3. **Combine the approximations**: Now we can rewrite the limit using these approximations: \[ \lim _{n \rightarrow \infty} \frac{\sqrt{4 n^{3}-n}}{2 n \sqrt{n-1}} \approx \lim _{n \rightarrow \infty} \frac{2n^{3/2}}{2n^{3/2}} = \lim _{n \rightarrow \infty} 1 = 1. \] Thus, the final result is: \[ \lim _{n \rightarrow \infty} \frac{\sqrt{4 n^{3}-n}}{2 n \sqrt{n-1}} = 1. \]