Skip to content
← All questions

Why is the assumption n>3n > \sqrt{3} necessary when solving the error inequality for the sequence 3n²/(n²−3)?

The assumption n>3n > \sqrt{3} ensures that the denominator n2−3n^2 - 3 is positive. This is crucial because when solving the inequality 9n2−3<ε\frac{9}{n^2 - 3} < \varepsilon, we multiply both sides by n2−3n^2 - 3. If the denominator were negative, the inequality sign would flip, leading to an incorrect solution. Since nn is a positive integer, n≥2n \ge 2 satisfies this condition, but the algebraic derivation explicitly states n>3n > \sqrt{3}.

Conditions

  • Solving 9n2−3<ε\frac{9}{n^2 - 3} < \varepsilon.
  • nn is a positive integer.
  • Denominator is n2−3n^2 - 3.

Reasoning, step by step

  1. Identify the error term 9n2−3\frac{9}{n^2 - 3}.
  2. Note that multiplying by the denominator requires knowing its sign.
  3. Assume n>3n > \sqrt{3} so n2−3>0n^2 - 3 > 0.
  4. Proceed to solve the inequality without flipping the sign.
  5. Verify that the resulting NN satisfies N>3N > \sqrt{3}.

Example

The script states: 'Assuming n > √3 to ensure the denominator is positive, the inequality is solved for n, resulting in n > √(9/ε+39/ε + 3).'

Common misconceptions

  • Forgetting to check the sign of the denominator before multiplying.
  • Believing the inequality holds for all nn without domain restrictions.

Watch the explanation

Connected concepts

Explore next

Related questions

Find a method

↗
Understand why

↗
Meet the concept

↗
Find a method

↗

Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.