The problem asks for the limit as $x$ approaches infinity of the function $\sqrt[3]{x^3 + x^2 + 1 - x}$.

Let's analyze the expression inside the cube root:

$f(x) = \sqrt[3]{x^3 + x^2 + 1 - x}$

Step 1: Simplify the expression inside the cube root

The dominant term inside the cube root as $x$ approaches infinity is $x^3$, because it's the highest degree term. Therefore, we can factor $x^3$ out of the expression inside the cube root:

$f(x) = \sqrt[3]{x^3(1 + \frac{1}{x} + \frac{1}{x^2} - \frac{1}{x^2})}$

Simplifying the expression inside the cube root:

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

As $x$ approaches infinity, the fraction terms $\frac{1}{x}$, $\frac{1}{x^2}$, and $\frac{1}{x^3}$ approach zero:

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

Step 2: Evaluate the limit

As $x$ approaches infinity:

$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} x = \infty$

Therefore, the limit of the given function as $x$ approaches infinity is $\infty$.

Would you like a more detailed explanation, or do you have any questions? Here are some related questions you might find interesting:

1. How do you evaluate limits at infinity involving polynomial expressions?
2. What happens to functions involving roots as $x$ approaches infinity?
3. How do you determine the dominant term in a polynomial?
4. Can limits at infinity ever result in a finite number?
5. How does L'Hôpital's Rule apply to limits involving infinity?

Tip: When finding limits involving infinity, focusing on the highest-degree term can often simplify the problem significantly.