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We are tasked with finding the limit:

\[
\lim_{x \to \infty} \sqrt[3]{x^3 - 4x^2 - x}
\]

### Step 1: Analyzing the Leading Term
First, let's examine the function \( x^3 - 4x^2 - x \) as \( x \to \infty \).

The dominant term is \( x^3 \), which suggests the cube root will behave approximately like \( \sqrt[3]{x^3} = x \) for large \( x \).

### Step 2: Simplification
We can factor \( x^3 \) from the expression inside the cube root:

\[
\sqrt[3]{x^3 - 4x^2 - x} = \sqrt[3]{x^3 \left( 1 - \frac{4}{x} - \frac{1}{x^2} \right)}
\]

### Step 3: Approximating as \( x \to \infty \)
As \( x \to \infty \), the terms \( \frac{4}{x} \) and \( \frac{1}{x^2} \) approach 0. Therefore, the expression inside the cube root tends to:

\[
\sqrt[3]{x^3 \cdot (1 - 0 - 0)} = \sqrt[3]{x^3} = x
\]

### Step 4: Conclusion
Thus, the limit of the expression as \( x \to \infty \) is:

\[
\lim_{x \to \infty} \sqrt[3]{x^3 - 4x^2 - x} = \infty
\]

### Final Answer:
\[
\boxed{\infty}
\]

Would you like further details or have any questions? 

Here are 5 related questions to deepen your understanding:
1. How would the limit change if the leading term were \( x^2 \) instead of \( x^3 \)?
2. Can you explain how factoring out the highest power of \( x \) helped simplify the problem?
3. What happens if we compute the same limit as \( x \to -\infty \)?
4. How would the limit change if there was a constant instead of a linear term inside the cube root?
5. How does L'Hopital's rule apply to problems where the result is indeterminate, like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \)?

**Tip:** When computing limits at infinity for polynomial or rational functions, always focus on the highest degree terms.

Compute the following limit using L'Hopital's rule if appropriate: \( \lim_{x \to \infty} \sqrt[3]{x^3 - 4x^2 - x} \)

Find the limit of a cubic polynomial under a cube root as x approaches infinity. This example simplifies the expression and analyzes the leading term. Suitable for college-level calculus students.

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