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Let's solve the given limit:

\[
\lim_{x \to 0} \frac{\sin(3x)}{\sqrt{5x + 4} - 2}
\]

### Step 1: Simplify the denominator

The denominator is \(\sqrt{5x + 4} - 2\), which approaches 0 as \(x\) approaches 0. To simplify, multiply the numerator and denominator by the conjugate of the denominator:

\[
\frac{\sin(3x)}{\sqrt{5x + 4} - 2} \cdot \frac{\sqrt{5x + 4} + 2}{\sqrt{5x + 4} + 2} = \frac{\sin(3x) \cdot (\sqrt{5x + 4} + 2)}{(5x + 4) - 4}
\]

This simplifies to:

\[
\frac{\sin(3x) \cdot (\sqrt{5x + 4} + 2)}{5x}
\]

### Step 2: Break down the limit expression

We can now rewrite the limit as:

\[
\lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot \frac{3x}{5x} \cdot (\sqrt{5x + 4} + 2)
\]

Since \(\lim_{x \to 0} \frac{\sin(3x)}{3x} = 1\) (a standard limit), and as \(x \to 0\), \(\sqrt{5x + 4} + 2 \to 4\), we can evaluate the limit:

\[
= 1 \cdot \frac{3}{5} \cdot 4 = \frac{12}{5}
\]

### Final Answer:
The correct answer is **b) \(\frac{12}{5}\)**.

Would you like more details on any step?

Here are 5 related questions for further practice:
1. What is the limit \(\lim_{x \to 0} \frac{\sin(5x)}{\sqrt{7x + 9} - 3}\)?
2. How do you simplify limits involving square roots and trigonometric functions?
3. What is L'Hôpital's Rule, and when can it be applied to limits?
4. How do conjugates help in limit simplifications?
5. Can limits always be evaluated using standard limit properties?

**Tip:** Always check if multiplying by the conjugate simplifies the expression when you encounter square roots in limits.

What is the value of the limit lim_{x->0} sin(3x) / (sqrt(5x+4)-2)?

Learn how to solve the limit problem involving trigonometric functions and square roots: lim_{x->0} (sin(3x) / (sqrt(5x+4)-2)). This solution breaks down the steps of using conjugates and limit properties for trigonometric functions. Suitable for advanced high school or early college students.

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