Let's analyze the given expression from the image. It appears that you're asked to find the following limit:

$\lim_{x \to \infty} \frac{\sqrt{x^4 + 9}}{x^2 - 2x - 3} \div \frac{x^2 + 9}{1}$

Step-by-step breakdown:

1. First, simplify the complex expression: $\frac{\sqrt{x^4 + 9}}{x^2 - 2x - 3} \div \frac{x^2 + 9}{1} = \frac{\sqrt{x^4 + 9}}{x^2 - 2x - 3} \times \frac{1}{x^2 + 9}$ This simplifies to: $\frac{\sqrt{x^4 + 9}}{(x^2 - 2x - 3)(x^2 + 9)}$

2. Now, analyze the behavior as $x \to \infty$:

   • The term $\sqrt{x^4 + 9}$ behaves like $x^2$ for large $x$, because $x^4$ dominates over the constant 9. So: $\sqrt{x^4 + 9} \approx x^2 \quad \text{as} \quad x \to \infty$

   • Both $x^2 - 2x - 3$ and $x^2 + 9$ behave like $x^2$ as well when $x$ becomes large, since higher powers dominate the expression.

3. Substitute these approximations into the limit expression: $\frac{x^2}{x^2 \cdot x^2} = \frac{x^2}{x^4} = \frac{1}{x^2}$ As $x \to \infty$, $\frac{1}{x^2} \to 0$.

Thus, the limit is: $\lim_{x \to \infty} \frac{\sqrt{x^4 + 9}}{(x^2 - 2x - 3)(x^2 + 9)} = 0$

Final Answer:

$\lim_{x \to \infty} \frac{\sqrt{x^4 + 9}}{(x^2 - 2x - 3)(x^2 + 9)} = 0$

Would you like more details or have any questions on this?
Here are 5 related questions to expand on this topic:

1. What is the general method for finding limits involving square roots and polynomials?
2. How do we approximate dominant terms for large $x$?
3. What is the behavior of rational functions at infinity?
4. How do you factor complicated polynomials to simplify limits?
5. Can L'Hopital's Rule be applied to similar limits, and how?

Tip: When dealing with limits as $x \to \infty$, always focus on the highest power of $x$ in both the numerator and denominator. This simplifies the analysis.