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Let's solve the limits step by step, and we will first determine if indeterminate forms exist. Here's the breakdown for each problem:

### a) \(\lim_{x \to 2} \frac{\sqrt{6 - x} - 2}{\sqrt{3 - x} - 1}\)

1. **Substitute \(x = 2\)** into the expression:
   \[
   \frac{\sqrt{6 - 2} - 2}{\sqrt{3 - 2} - 1} = \frac{\sqrt{4} - 2}{\sqrt{1} - 1} = \frac{2 - 2}{1 - 1} = \frac{0}{0}
   \]
   This is an indeterminate form, so we need to simplify the expression using algebraic techniques.

2. **Multiply by the conjugates** of both the numerator and denominator:
   \[
   \frac{\sqrt{6 - x} - 2}{\sqrt{3 - x} - 1} \times \frac{\sqrt{6 - x} + 2}{\sqrt{3 - x} + 1} = \frac{(6 - x) - 4}{(3 - x) - 1}
   \]
   Simplifying further:
   \[
   \frac{6 - x - 4}{3 - x - 1} = \frac{2 - x}{2 - x}
   \]
   The expression simplifies to \(1\), as long as \(x \neq 2\).

3. **Evaluate the limit**:
   \[
   \lim_{x \to 2} 1 = 1
   \]

Thus, the result is:
\[
\lim_{x \to 2} \frac{\sqrt{6 - x} - 2}{\sqrt{3 - x} - 1} = 1
\]

---

### b) \(\lim_{x \to -4} \frac{x^2 + 5x + 4}{x^2 + 3x - 4}\)

1. **Substitute \(x = -4\)**:
   \[
   \frac{(-4)^2 + 5(-4) + 4}{(-4)^2 + 3(-4) - 4} = \frac{16 - 20 + 4}{16 - 12 - 4} = \frac{0}{0}
   \]
   This is an indeterminate form, so we must factor the expression.

2. **Factor the numerator and denominator**:
   \[
   \frac{(x + 4)(x + 1)}{(x + 4)(x - 1)}
   \]
   Cancel the common factor \((x + 4)\):
   \[
   \frac{x + 1}{x - 1}
   \]

3. **Evaluate the limit**:
   \[
   \lim_{x \to -4} \frac{x + 1}{x - 1} = \frac{-4 + 1}{-4 - 1} = \frac{-3}{-5} = \frac{3}{5}
   \]

Thus, the result is:
\[
\lim_{x \to -4} \frac{x^2 + 5x + 4}{x^2 + 3x - 4} = \frac{3}{5}
\]

---

### c) \(\lim_{u \to \infty} \frac{4u^4 + 5}{(u^2 - 2)(2u^2 - 1)}\)

1. **Analyze the degrees of the polynomial**:
   - The numerator is \(4u^4 + 5\), and the denominator is \((u^2 - 2)(2u^2 - 1)\).
   - The leading term in the numerator is \(4u^4\).
   - Expanding the denominator:
     \[
     (u^2 - 2)(2u^2 - 1) = 2u^4 - u^2 - 4u^2 + 2 = 2u^4 - 5u^2 + 2
     \]
   - The leading term in the denominator is \(2u^4\).

2. **Evaluate the limit of the leading terms**:
   \[
   \lim_{u \to \infty} \frac{4u^4}{2u^4} = \frac{4}{2} = 2
   \]

Thus, the result is:
\[
\lim_{u \to \infty} \frac{4u^4 + 5}{(u^2 - 2)(2u^2 - 1)} = 2
\]

---

### d) \(\lim_{x \to \infty} \frac{\sqrt{9x^6 - x}}{x^3 + 1}\)

1. **Divide both numerator and denominator by \(x^3\)** (the highest power of \(x\) in the denominator):
   \[
   \frac{\sqrt{9x^6 - x}}{x^3 + 1} = \frac{x^3 \sqrt{9 - \frac{1}{x^5}}}{x^3 \left(1 + \frac{1}{x^3}\right)}
   \]

2. **Simplify as \(x \to \infty\)**:
   - In the numerator, \(\sqrt{9 - \frac{1}{x^5}} \to 3\).
   - In the denominator, \(1 + \frac{1}{x^3} \to 1\).

Thus, the limit becomes:
\[
\lim_{x \to \infty} \frac{3x^3}{x^3} = 3
\]

Thus, the result is:
\[
\lim_{x \to \infty} \frac{\sqrt{9x^6 - x}}{x^3 + 1} = 3
\]

---

If you have any questions about these solutions or need further explanations, feel free to ask!

### Relative questions:
1. What are the common methods for handling limits involving indeterminate forms?
2. How do conjugates help in simplifying limits involving square roots?
3. Why is factoring important in resolving polynomial indeterminate limits?
4. How does the behavior of polynomials at infinity affect limit evaluations?
5. What techniques can be used for handling limits that go to infinity?

### Tip:
Always check for indeterminate forms before attempting to simplify a limit problem.

Realice los siguientes límites, pero para ello primero defina si existe una indeterminación.

Explore detailed step-by-step solutions to various limit problems involving indeterminate forms. This guide includes techniques like conjugate multiplication, polynomial factorization, and analysis of behavior at infinity. Suitable for university-level students studying calculus.

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