Let's analyze and **bound the numerator and denominator** for the given expression: \[ \left| \frac{4n - 3}{3(3n^2 - n)} \right| \] ### **Numerator: \(4n - 3\)** - For large \(n\), \(4n\) dominates over \(-3\). - So, \(4n - 3 \approx 4n\). - **Lower bound:** \(4n - 3 > 0\) for \(n > 0.75\), so for \(n \geq 1\), numerator is positive. - **Upper bound:** \(4n - 3 < 4n\). ### **Denominator: \(3(3n^2 - n)\)** - For large \(n\), \(3n^2\) dominates over \(-n\). - So, \(3(3n^2 - n) \approx 9n^2\). - **Lower bound:** \(3n^2 - n > 0\) for \(n > 0\), so for \(n \geq 1\), denominator is positive. - **Upper bound:** \(3n^2 - n < 3n^2\) for \(n > 0\). ### **Absolute Value** Since \(n > 1\), both numerator and denominator are positive, so the absolute value is just the fraction. ### **Bounding the Expression** For \(n \geq 1\): - **Numerator:** \(4n - 3 < 4n\) - **Denominator:** \(3(3n^2 - n) > 0\), and \(3(3n^2 - n) < 9n^2\) So, \[ \left| \frac{4n - 3}{3(3n^2 - n)} \right| < \frac{4n}{9n^2} = \frac{4}{9n} \] For a **lower bound** (for \(n \geq 1\)): - \(4n - 3 > n\) for \(n \geq 2\) (since \(4n - 3 - n = 3n - 3 > 0\) for \(n > 1\)) - \(3(3n^2 - n) > 6n^2\) for \(n \geq 1\) (since \(3n^2 - n > 2n^2\) for \(n > 1\)) Thus, \[ \left| \frac{4n - 3}{3(3n^2 - n)} \right| > \frac{n}{6n^2} = \frac{1}{6n} \] ### **Summary** For large \(n\) (say \(n \geq 2\)): \[ \frac{1}{6n} < \left| \frac{4n - 3}{3(3n^2 - n)} \right| < \frac{4}{9n} \] Let me know if you'd like a more precise bound or for a different range of \(n\)!