The expression given in the image is:

$\lim_{n \to \infty} \frac{n^2 + 4}{9 - 3\left(\frac{n^2 + 1}{n^3 + 5n}\right) + 2}$

Step-by-Step Solution:

Step 1: Simplify the denominator

We begin by simplifying the expression in the denominator. Notice the complex term involving $n^2 + 1$ and $n^3 + 5n$.

$\frac{n^2 + 1}{n^3 + 5n} = \frac{n^2(1 + \frac{1}{n^2})}{n^3(1 + \frac{5}{n^2})}$

As $n \to \infty$, both $\frac{1}{n^2} \to 0$ and $\frac{5}{n^2} \to 0$, so the term simplifies to:

$\frac{n^2 + 1}{n^3 + 5n} \sim \frac{n^2}{n^3} = \frac{1}{n}$

Thus, the expression in the denominator becomes approximately:

$9 - 3 \cdot \frac{1}{n} + 2 \quad \text{as} \quad n \to \infty$

Simplifying:

$9 - 3\left(\frac{1}{n}\right) + 2 = 11 - \frac{3}{n}$

Step 2: Take the limit as $n \to \infty$

As $n \to \infty$, the term $\frac{3}{n} \to 0$. So the denominator simplifies to:

$11$

Step 3: Analyze the numerator

The numerator is $n^2 + 4$. As $n \to \infty$, this term behaves like $n^2$.

Step 4: Combine the results

The entire expression for the limit becomes:

$\lim_{n \to \infty} \frac{n^2 + 4}{11}$

As $n \to \infty$, this simplifies to:

$\lim_{n \to \infty} \frac{n^2}{11} = \infty$

Thus, the limit is:

$\boxed{\infty}$

Would you like further details or have any specific questions about this solution?

Here are some related questions to explore:

1. What happens if the degree of the numerator and denominator are the same?
2. How does the behavior of rational functions change as $n \to \infty$?
3. How would the limit change if the term in the numerator were $n$ instead of $n^2$?
4. Can L'Hopital's Rule be applied to this limit? Why or why not?
5. What if the highest-degree term in the denominator had been $n^4$ instead of $n^3$?

Tip: When working with limits at infinity, comparing the degrees of terms in the numerator and denominator can help you quickly identify the limit behavior.