We are given the limit problem:

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

Step-by-Step Solution:

1. Substitute $x = 0$: First, let's substitute $x = 0$ directly into the expression to check if it's an indeterminate form.

   $\frac{\sin(3(0))}{\sqrt{5(0) + 4} - 2} = \frac{\sin(0)}{\sqrt{4} - 2} = \frac{0}{2 - 2} = \frac{0}{0}$

   Since we get the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule, which states that if a limit is in the indeterminate form $\frac{0}{0}$, we can differentiate the numerator and denominator and then evaluate the limit.

2. Differentiate the numerator: The derivative of $\sin(3x)$ is: $\frac{d}{dx}[\sin(3x)] = 3\cos(3x)$

3. Differentiate the denominator: The denominator is $\sqrt{5x + 4} - 2$. The derivative of $\sqrt{5x + 4}$ is: $\frac{d}{dx}[\sqrt{5x + 4}] = \frac{5}{2\sqrt{5x + 4}}$ Since the derivative of $-2$ is $0$, the derivative of the entire denominator is: $\frac{5}{2\sqrt{5x + 4}}$

4. Set up the new limit: After applying L'Hopital's Rule, we now have: $\lim_{x \to 0} \frac{3\cos(3x)}{\frac{5}{2\sqrt{5x + 4}}}$

5. Simplify the expression: Substituting $x = 0$ in the simplified expression: $\lim_{x \to 0} \frac{3\cos(3x)}{\frac{5}{2\sqrt{4}}} = \frac{3 \cdot \cos(0)}{\frac{5}{2 \cdot 2}} = \frac{3 \cdot 1}{\frac{5}{4}} = 3 \cdot \frac{4}{5} = \frac{12}{5}$

   Hence, the value of the limit is: $\frac{12}{5}$

Answer:

The correct option is (b) $\frac{12}{5}$.

Do you have any questions or would you like more details on any step?

Here are some related questions for practice:

1. What is L'Hopital's Rule, and when is it applicable?
2. How do you find the derivative of a square root function?
3. What are the standard limits involving trigonometric functions like $\sin(x)$ and $\cos(x)$?
4. How do you simplify complex fractions?
5. Can L'Hopital's Rule be applied multiple times if necessary?

Tip: When applying L'Hopital's Rule, always check that the limit is initially in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$.