12–24. Limits of sequences Evaluate the limit of the sequence or state that it does not exist.
aₙ = ((3n² + 2n + 1) · sin(n)) / (4n³ + n) (Hint: Use the Squeeze Theorem.)

Verified step by step guidance

1. Identify the given sequence: \(a_n = \frac{(3n^2 + 2n + 1) \cdot \sin(n)}{4n^3 + n}\).
2. Analyze the behavior of the numerator and denominator separately as \(n\) approaches infinity. The numerator grows roughly like \$3n^2\( times \(\sin(n)\), and the denominator grows like \)4n^3$.
3. Note that \(\sin(n)\) oscillates between \(-1\) and \$1\(, so \(-1 \leq \sin(n) \leq 1\). Use this to create inequalities for \)a_n$:
4. \[-\frac{3n^2 + 2n + 1}{4n^3 + n} \leq a_n \leq \frac{3n^2 + 2n + 1}{4n^3 + n}.\]
5. Simplify the bounding expressions by dividing numerator and denominator by \(n^3\) to understand their limits:
6. \[-\frac{3/n + 2/n^2 + 1/n^3}{4 + 1/n^2} \leq a_n \leq \frac{3/n + 2/n^2 + 1/n^3}{4 + 1/n^2}.\]
7. Apply the Squeeze Theorem: since both bounds approach \$0$ as \(n \to \infty\), conclude that \(\lim_{n \to \infty} a_n = 0\).

Key Concepts

Limits of Sequences

The limit of a sequence describes the value that the terms of the sequence approach as the index n goes to infinity. If the terms get arbitrarily close to a specific number, the sequence converges to that limit; otherwise, it diverges.

Squeeze Theorem

The Squeeze Theorem helps find limits by 'trapping' a sequence between two others that have the same limit. If aₙ ≤ bₙ ≤ cₙ for all n beyond some point, and both aₙ and cₙ converge to L, then bₙ also converges to L.

Behavior of Trigonometric Functions in Limits

Trigonometric functions like sin(n) oscillate between -1 and 1 and do not have limits as n approaches infinity. When combined with sequences that tend to zero, their bounded nature allows the use of the Squeeze Theorem to evaluate the overall limit.