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To solve the given limit:

\[
\lim_{x \to \infty} \frac{4(\ln x)^3}{x}
\]

we can apply L'Hôpital's Rule because it is an indeterminate form of type \(\frac{\infty}{\infty}\).

### Step-by-Step Solution:

1. **Apply L'Hôpital's Rule**:
   L'Hôpital's Rule states that if the limit is in an indeterminate form, we can take the derivative of the numerator and the denominator:
   
   \[
   \lim_{x \to \infty} \frac{4(\ln x)^3}{x} = \lim_{x \to \infty} \frac{\frac{d}{dx} \left[ 4(\ln x)^3 \right]}{\frac{d}{dx} [x]}
   \]

2. **Differentiate the numerator and denominator**:
   
   - Derivative of the numerator: 
     
     \[
     \frac{d}{dx} \left[ 4(\ln x)^3 \right] = 4 \cdot 3(\ln x)^2 \cdot \frac{1}{x} = \frac{12(\ln x)^2}{x}
     \]
   
   - Derivative of the denominator:
     
     \[
     \frac{d}{dx} [x] = 1
     \]

   So the limit becomes:
   
   \[
   \lim_{x \to \infty} \frac{\frac{12(\ln x)^2}{x}}{1} = \lim_{x \to \infty} \frac{12(\ln x)^2}{x}
   \]

3. **Apply L'Hôpital's Rule again** because the limit is still of the form \(\frac{\infty}{\infty}\):

   \[
   \lim_{x \to \infty} \frac{12(\ln x)^2}{x} = \lim_{x \to \infty} \frac{\frac{d}{dx} \left[ 12(\ln x)^2 \right]}{\frac{d}{dx} [x]}
   \]

4. **Differentiate the numerator again**:
   
   - Derivative of \(12(\ln x)^2\):
     
     \[
     \frac{d}{dx} \left[ 12(\ln x)^2 \right] = 12 \cdot 2(\ln x) \cdot \frac{1}{x} = \frac{24 \ln x}{x}
     \]

   So the limit becomes:
   
   \[
   \lim_{x \to \infty} \frac{\frac{24 \ln x}{x}}{1} = \lim_{x \to \infty} \frac{24 \ln x}{x}
   \]

5. **Apply L'Hôpital's Rule one more time**:
   
   \[
   \lim_{x \to \infty} \frac{24 \ln x}{x} = \lim_{x \to \infty} \frac{\frac{d}{dx} [24 \ln x]}{\frac{d}{dx} [x]}
   \]
   
   - Derivative of \(24 \ln x\):
     
     \[
     \frac{d}{dx} [24 \ln x] = \frac{24}{x}
     \]

   So the limit is:
   
   \[
   \lim_{x \to \infty} \frac{\frac{24}{x}}{1} = \lim_{x \to \infty} \frac{24}{x} = 0
   \]

### Final Answer:

\[
\lim_{x \to \infty} \frac{4(\ln x)^3}{x} = 0
\]

Would you like more details or have any questions?

### Related Questions:
1. What is L'Hôpital's Rule, and when can it be applied?
2. How would the solution change if the numerator was \( (\ln x)^4 \) instead?
3. What are other indeterminate forms that L'Hôpital's Rule can address?
4. How does the growth rate of \( \ln x \) compare to polynomials or exponential functions as \( x \to \infty \)?
5. Can the limit be solved using a method other than L'Hôpital’s Rule?

### Tip:
Remember, L'Hôpital's Rule can only be applied to limits that are in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).

Find the limit \( \lim_{x \to \infty} \frac{4(\ln x)^3}{x} \) using L'Hôpital's Rule.

Learn to find the limit of \( \lim_{x \to \infty} \frac{4(\ln x)^3}{x} \) by applying L'Hôpital's Rule. This detailed solution includes step-by-step differentiation and multiple applications of the rule, perfect for calculus students tackling indeterminate forms.

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