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We are tasked with solving the following limit:

\[
\lim_{x \to \infty} x\left( \sqrt{x^2 + 2x} - 2\sqrt{x^2 + x + x} \right)
\]

### Step-by-Step Solution:

#### 1. Simplifying the square root terms:
For large \( x \), we can factor \( x^2 \) out of each square root to simplify them.

- In \( \sqrt{x^2 + 2x} \), factor \( x^2 \) out:
  \[
  \sqrt{x^2 + 2x} = \sqrt{x^2(1 + \frac{2}{x})} = x\sqrt{1 + \frac{2}{x}}
  \]
  
- Similarly, for \( 2\sqrt{x^2 + 2x} \):
  \[
  2\sqrt{x^2 + 2x} = 2x\sqrt{1 + \frac{2}{x}}
  \]

#### 2. Applying the limit:
As \( x \to \infty \), the terms \( \frac{2}{x} \to 0 \), so we can approximate:

- \( \sqrt{1 + \frac{2}{x}} \to 1 \)
- \( 2\sqrt{1 + \frac{2}{x}} \to 2 \)

Thus, for large \( x \), the expression inside the limit becomes approximately:

\[
x \left( x(1 + \frac{2}{x}) - 2x\left( \) = 
\--- Placeholder

lim_{x→∞} x(√(x² + 2x) - 2√(x² + x + x))

Explore how to solve the limit lim_{x→∞} x(√(x² + 2x) - 2√(x² + x + x)) by applying square root simplification and limit laws. Detailed steps provided. Suitable for advanced high school or early university students.

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