The expression you provided is:

$\lim_{{x \to 0}} \frac{(x^2 - 1) \sin(6x)}{x^3 + 3x^2 + 2x}$

Step-by-Step Solution:

1. Evaluate the limit directly: Plugging in $x = 0$ directly, we get: $\frac{(0^2 - 1) \sin(6(0))}{0^3 + 3(0^2) + 2(0)} = \frac{-1 \cdot 0}{0} = \frac{0}{0}$ which results in the indeterminate form $\frac{0}{0}$. Thus, we can apply L'Hôpital's Rule.

2. L'Hôpital's Rule: This rule states that for indeterminate forms like $\frac{0}{0}$, we can take the derivative of the numerator and the denominator separately and then evaluate the limit.

   • Numerator's derivative: $\frac{d}{dx} \left( (x^2 - 1) \sin(6x) \right)$ Use the product rule: $\frac{d}{dx} \left( (x^2 - 1) \right) \cdot \sin(6x) + (x^2 - 1) \cdot \frac{d}{dx} \left( \sin(6x) \right)$ $= (2x) \sin(6x) + (x^2 - 1) \cdot 6 \cos(6x)$

   • Denominator's derivative: $\frac{d}{dx} \left( x^3 + 3x^2 + 2x \right) = 3x^2 + 6x + 2$

3. New limit: Now, we substitute the derivatives back into the limit expression: $\lim_{{x \to 0}} \frac{2x \sin(6x) + (x^2 - 1) \cdot 6 \cos(6x)}{3x^2 + 6x + 2}$ Let's evaluate this at $x = 0$:

   • Numerator: When $x = 0$, $2(0) \sin(6(0)) + (0^2 - 1) \cdot 6 \cos(6(0)) = 0 - 1 \cdot 6 = -6$
   • Denominator: When $x = 0$, $3(0^2) + 6(0) + 2 = 2$

   Therefore, the limit becomes: $\frac{-6}{2} = -3$

Thus, the value of the limit is:

$\boxed{-3}$

Would you like more details on any step or related questions? Here are five related questions you might find interesting:

1. How does L'Hôpital's Rule work for other indeterminate forms like $\frac{\infty}{\infty}$?
2. Can L'Hôpital's Rule always be applied to indeterminate forms?
3. What are some common cases where trigonometric limits are used?
4. How does the product rule work when differentiating functions like $(x^2 - 1)\sin(6x)$?
5. What happens if the function is not in an indeterminate form? How do we solve limits then?

Tip: Always check if a limit leads to an indeterminate form before deciding on applying L'Hôpital’s Rule.