1. **State the problem:** Compute the limit $$\lim_{x \to \infty} \sqrt[3]{x^3 - 7x^2 - x}$$ using L'Hospital's rule if appropriate. 2. **Analyze the expression:** As $$x \to \infty$$, inside the cube root we have $$x^3 - 7x^2 - x$$. The dominant term is $$x^3$$. 3. **Simplify the expression:** We can factor out $$x^3$$ inside the cube root: $$\sqrt[3]{x^3 - 7x^2 - x} = \sqrt[3]{x^3\left(1 - \frac{7}{x} - \frac{1}{x^2}\right)} = \sqrt[3]{x^3} \cdot \sqrt[3]{1 - \frac{7}{x} - \frac{1}{x^2}}$$ 4. **Simplify further:** The cube root of $$x^3$$ is $$x$$, so $$ = x \cdot \sqrt[3]{1 - \frac{7}{x} - \frac{1}{x^2}}$$ 5. **Evaluate the limit:** As $$x \to \infty$$, the fractions $$\frac{7}{x} \to 0$$ and $$\frac{1}{x^2} \to 0$$, so $$\sqrt[3]{1 - \frac{7}{x} - \frac{1}{x^2}} \to \sqrt[3]{1} = 1$$ 6. **Therefore:** $$\lim_{x \to \infty} \sqrt[3]{x^3 - 7x^2 - x} = \lim_{x \to \infty} x \cdot 1 = \infty$$ 7. **L'Hospital's rule applicability:** The expression is not an indeterminate form of quotient type, and the limit can be found by simplification without L'Hospital's rule. **Final answer:** $$\infty$$