The given expression asks to evaluate the limit as $x \to \infty$ of the following function:

$\lim_{x \to \infty} \sqrt{\frac{(3x^3 + 5x - 1)^3 (9 - 2x^2)^2}{(2x + 7) (4x^4 - 5)^3}}$

Step-by-step breakdown:

1. Analyze the asymptotic behavior:

   • As $x \to \infty$, we focus on the highest degree terms in both the numerator and denominator, since those will dominate the behavior of the expression.

   • Numerator:

     • $3x^3 + 5x - 1 \sim 3x^3$
     • $9 - 2x^2 \sim -2x^2$ Therefore, the numerator behaves like: $(3x^3)^3 (-2x^2)^2 = 27x^9 \cdot 4x^4 = 108x^{13}$

   • Denominator:

     • $2x + 7 \sim 2x$
     • $4x^4 - 5 \sim 4x^4$ Therefore, the denominator behaves like: $2x \cdot (4x^4)^3 = 2x \cdot 64x^{12} = 128x^{13}$

2. Simplify the asymptotic limit: So, the expression inside the square root becomes asymptotically: $\frac{108x^{13}}{128x^{13}} = \frac{108}{128} = \frac{27}{32}$

3. Take the square root: $\sqrt{\frac{27}{32}} = \frac{\sqrt{27}}{\sqrt{32}} = \frac{3\sqrt{3}}{4\sqrt{2}} = \frac{3\sqrt{6}}{8}$

Thus, the value of the limit is:

$\boxed{\frac{3\sqrt{6}}{8}}$

Let me know if you would like further details on any of the steps!

Related questions:

1. How would the solution change if the degree of the polynomial in the numerator was higher?
2. What happens to the limit if the denominator had an additional higher-degree term?
3. How can you determine the dominant term for complex functions as $x \to \infty$?
4. Can L'Hopital's Rule be applied to this type of limit problem?
5. What is the behavior of this function as $x \to -\infty$?

Tip:

When evaluating limits at infinity, always focus on the highest power terms, as they dominate the behavior of the function.