The given limit problem is:

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

To solve this, let's simplify the expression inside the limit step by step.

Step 1: Approximate the square roots for large $x$

When $x \to \infty$, the dominant term inside the square roots is $x^2$. Thus, we can factor $x^2$ from the terms inside the square roots:

$\sqrt{x^2 + 2x} = x \sqrt{1 + \frac{2}{x}} \quad \text{and} \quad \sqrt{x^2 + x} = x \sqrt{1 + \frac{1}{x}}$

For large $x$, we can use the first two terms of the binomial expansion for square roots:

$\sqrt{1 + \frac{2}{x}} \approx 1 + \frac{1}{x} \quad \text{and} \quad \sqrt{1 + \frac{1}{x}} \approx 1 + \frac{1}{2x}$

Thus:

$\sqrt{x^2 + 2x} \approx x \left( 1 + \frac{1}{x} \right) = x + 1$ $\sqrt{x^2 + x} \approx x \left( 1 + \frac{1}{2x} \right) = x + \frac{1}{2}$

Step 2: Substitute into the expression

Now substitute these approximations into the limit expression:

$x \left( (x + 1) - 2(x + \frac{1}{2}) \right)$

Simplify the expression inside the parentheses:

$(x + 1) - 2(x + \frac{1}{2}) = x + 1 - 2x - 1 = -x$

Thus, the expression becomes:

$x \cdot (-x) = -x^2$

Step 3: Evaluate the limit

The limit now becomes:

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

As $x \to \infty$, this tends to $-\infty$.

Final Answer:

$\boxed{-\infty}$

Would you like more details on any specific step? Here are 5 related questions to further explore this concept:

1. How does binomial expansion help in simplifying square roots?
2. What happens if we ignore the second terms in the approximations for the square roots?
3. How can this method be generalized for similar limits involving square roots?
4. What is the role of factoring the dominant term when simplifying limits?
5. Could L'Hopital's Rule be used as an alternative approach here?

Tip: When simplifying square roots for large $x$, factor out the highest power of $x$ inside the square root and approximate using binomial expansion.