The given problem is:

$\lim_{{x \to 0}} \frac{\sin(x) - x}{4x^3}$

Solution:

1. Substitute $x = 0$ into the limit to see if it's indeterminate: $\frac{\sin(0) - 0}{4(0)^3} = \frac{0 - 0}{0} = \frac{0}{0}$ This is an indeterminate form $\frac{0}{0}$, so we can apply L'Hôpital's Rule.

2. Apply L'Hôpital's Rule: Since the limit is in the form $\frac{0}{0}$, we differentiate the numerator and denominator with respect to $x$.

   • Numerator: $\frac{d}{dx}[\sin(x) - x] = \cos(x) - 1$
   • Denominator: $\frac{d}{dx}[4x^3] = 12x^2$

   Now, the limit becomes: $\lim_{{x \to 0}} \frac{\cos(x) - 1}{12x^2}$

3. Substitute $x = 0$ again: $\frac{\cos(0) - 1}{12(0)^2} = \frac{1 - 1}{0} = \frac{0}{0}$ We still have an indeterminate form, so we apply L'Hôpital's Rule again.

4. Differentiate the numerator and denominator again:

   • Numerator: $\frac{d}{dx}[\cos(x) - 1] = -\sin(x)$
   • Denominator: $\frac{d}{dx}[12x^2] = 24x$

   The limit is now: $\lim_{{x \to 0}} \frac{-\sin(x)}{24x}$

5. Substitute $x = 0$ again: $\frac{-\sin(0)}{24(0)} = \frac{0}{0}$ Another indeterminate form, so apply L'Hôpital's Rule once more.

6. Differentiate the numerator and denominator one last time:

   • Numerator: $\frac{d}{dx}[-\sin(x)] = -\cos(x)$
   • Denominator: $\frac{d}{dx}[24x] = 24$

   The limit becomes: $\lim_{{x \to 0}} \frac{-\cos(x)}{24} = \frac{-\cos(0)}{24} = \frac{-1}{24}$

Final Answer:

$\lim_{{x \to 0}} \frac{\sin(x) - x}{4x^3} = \frac{-1}{24}$

Would you like any further clarification or details? Here are 5 related questions:

1. How do you recognize when to apply L'Hôpital's Rule?
2. What are the conditions for applying L'Hôpital's Rule?
3. Can limits like this be solved without L'Hôpital's Rule?
4. What is the geometric interpretation of limits involving trigonometric functions?
5. How does Taylor expansion relate to solving limits?

Tip: If L'Hôpital's Rule gives the same indeterminate form repeatedly, consider checking higher-order derivatives or using series expansion to simplify the function around the point.