The given problem is asking for the limit of the function as $x \to \infty$:

$\lim_{x \to \infty} \sqrt{x^2 - 2x - x}$

First, let's simplify the expression inside the square root:

$\sqrt{x^2 - 2x - x} = \sqrt{x^2 - 3x}$

Now, factor the highest power of $x$ inside the square root:

$\sqrt{x^2 - 3x} = \sqrt{x^2(1 - \frac{3}{x})} = x\sqrt{1 - \frac{3}{x}}$

As $x \to \infty$, the term $\frac{3}{x}$ approaches 0. Therefore, the expression inside the square root simplifies to:

$x \sqrt{1 - 0} = x \times 1 = x$

Thus, the limit is:

$\lim_{x \to \infty} x = \infty$

Therefore, the correct answer is $\infty$, or option E.

Would you like more details or have any questions? Here are five related questions to deepen understanding:

1. How do we handle limits involving square roots as $x$ approaches infinity?
2. What happens to terms like $\frac{3}{x}$ as $x \to \infty$?
3. How does factoring out $x^2$ help simplify the limit?
4. Can you explain why the square root of a squared term yields positive results?
5. How can you apply the same method to solve similar limits involving polynomials?

Tip: When evaluating limits at infinity, focus on the highest power of $x$ to simplify the function.