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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.

Find the limit of \( \lim_{x \to \infty} \sqrt{x^2 - 2x - x} \).

Solve the limit problem \( \lim_{x \to \infty} \sqrt{x^2 - 2x - x} \). Learn how to simplify expressions under a square root and analyze the behavior of polynomials at infinity. Suitable for high school students in Grades 10-12.

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