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.