The given problem is:

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

Steps to solve:

1. Simplifying the square roots: For large $x$, we can factor out $x^2$ from inside the square roots and simplify the terms.

   • $\sqrt{x^2 + 2x} = \sqrt{x^2(1 + \frac{2}{x})} = x\sqrt{1 + \frac{2}{x}}$
   • $\sqrt{x^2 + x} = \sqrt{x^2(1 + \frac{1}{x})} = x\sqrt{1 + \frac{1}{x}}$

2. Substitute these into the original expression:

   = \lim_{{x \to \infty}} x^2 \left( \sqrt{1 + \frac{2}{x}} - 2\sqrt{1 + \frac{1}{x}} \right)$$

3. Expanding the square roots: For large $x$, we can use the approximation $\sqrt{1 + a} \approx 1 + \frac{a}{2}$ for small $a$.

   • $\sqrt{1 + \frac{2}{x}} \approx 1 + \frac{1}{x}$
   • $\sqrt{1 + \frac{1}{x}} \approx 1 + \frac{1}{2x}$

   Now substitute these approximations into the expression: $\lim_{{x \to \infty}} x^2 \left( \left(1 + \frac{1}{x}\right) - 2\left(1 + \frac{1}{2x}\right) \right)$ Simplifying the terms inside the parentheses: $\left( 1 + \frac{1}{x} \right) - 2\left( 1 + \frac{1}{2x} \right) = 1 + \frac{1}{x} - 2 - \frac{1}{x} = -1$

4. Final limit: Now the expression becomes: $\lim_{{x \to \infty}} x^2(-1) = -x^2$ As $x \to \infty$, this clearly tends to $-\infty$.

Conclusion:

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

Would you like more details or clarifications on this?

Here are five related questions for further exploration:

1. How would the result change if the coefficient in front of the second square root was different?
2. What happens if we change the powers inside the square roots, for example, replacing $x^2 + 2x$ with $x^3 + 2x^2$?
3. Can you explain how to derive the binomial approximation $\sqrt{1 + a} \approx 1 + \frac{a}{2}$?
4. What is the importance of considering the dominant term when evaluating limits as $x \to \infty$?
5. How would this problem differ if the limit was as $x \to 0$ instead of $x \to \infty$?

Tip: When simplifying square roots at infinity, factoring out the dominant term (usually the highest power of $x$) is key to revealing the leading behavior of the function.