We are tasked to find the limit of the sequence $\{a_n b_n\}$, where $\{a_n\}$ and $\{b_n\}$ are defined as follows:

1. The $n$th term of $\{a_n\}$ is: $a_n = \frac{2n^2 + 3n - \cos(n)}{2 - n^2}.$

2. The sequence $\{b_n\}$ is defined by: $b_n = a_n^2 + 2a_n - 5.$

To find $\lim_{n \to \infty} a_n b_n$, let's analyze step by step:

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

The numerator of $a_n$ is $2n^2 + 3n - \cos(n)$, and the denominator is $2 - n^2$. For large $n$, the leading term dominates. So: $a_n \sim \frac{2n^2}{-n^2} = -2 \quad \text{as } n \to \infty.$

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

Substitute $a_n \sim -2$ into the expression for $b_n$: $b_n = a_n^2 + 2a_n - 5.$ If $a_n \to -2$, then: $b_n \to (-2)^2 + 2(-2) - 5 = 4 - 4 - 5 = -5.$

Step 3: Asymptotic behavior of $a_n b_n$ as $n \to \infty$

Finally, consider the product: $a_n b_n \sim (-2)(-5) = 10.$

Thus: $\lim_{n \to \infty} a_n b_n = 10.$

Final Answer:

$\boxed{10}$

Would you like me to provide more details about any step or explore a related topic?

Expanding Questions:

1. Would you like an explanation for how $a_n$ was approximated?
2. Should I go into more detail about why higher-order terms of $-\cos(n)$ are negligible?
3. Do you want to verify the substitution $b_n = a_n^2 + 2a_n - 5$?
4. Should we compute numerical terms of $\{a_n b_n\}$ to confirm the limit?
5. Would you like a visual plot of $a_n b_n$ to illustrate its convergence?

Tip:

When finding limits of sequences involving rational expressions, always focus on the dominant terms of the numerator and denominator for large $n$.