Let us analyze and solve the problem systematically.

Problem Breakdown:

1. Sequence $\{a_n\}$ is defined as: $a_n = \frac{n^2 + 3n - \sin(n)}{2 - n^2}.$
2. Sequence $\{b_n\}$ is defined as: $b_n = a_n^2 + 2a_n - 5.$
3. We need to find the limit of the sequence $\{a_n b_n\}$ as $n \to \infty$.

Step 1: Analyze the behavior of $a_n$ as $n \to \infty$

• In the numerator: n^2 + 3n - \sin(n) \sim n^2 + 3n \quad \text{(since \sin(n) is bounded)}.
• In the denominator: 2 - n^2 \sim -n^2 \quad \text{(as n^2 dominates 2)}. Thus, the dominant behavior of $a_n$ is: $a_n \sim \frac{n^2}{-n^2} = -1.$ Hence, $\lim_{n \to \infty} a_n = -1$.

Step 2: Analyze the behavior of $b_n$ as $n \to \infty$

Using the definition of $b_n$: $b_n = a_n^2 + 2a_n - 5.$ Substitute $\lim_{n \to \infty} a_n = -1$: $b_n \to (-1)^2 + 2(-1) - 5 = 1 - 2 - 5 = -6.$ Thus, $\lim_{n \to \infty} b_n = -6$.

Step 3: Find the limit of $a_n b_n$

Using the fact that $\lim_{n \to \infty} a_n = -1$ and $\lim_{n \to \infty} b_n = -6$: $\lim_{n \to \infty} (a_n b_n) = (\lim_{n \to \infty} a_n)(\lim_{n \to \infty} b_n) = (-1)(-6) = 6.$

Final Answer:

$\boxed{6}$

Would you like me to provide further explanation?

Related Questions:

1. How does bounded behavior like $\sin(n)$ affect limits of sequences?
2. What are the common techniques for simplifying rational expressions in sequences?
3. What happens if the sequence $\{a_n\}$ diverges instead of converging?
4. Can this method be extended to sequences involving higher-degree polynomials?
5. How does the product rule for sequence limits work in general?

Tip:

When analyzing limits of rational expressions in sequences, focus on the dominant terms (highest powers) in the numerator and denominator for simplification.