The expression in the image is:

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

Let’s solve this step by step.

Step 1: Simplify the square root terms as $x \to +\infty$

For large $x$, we approximate the square roots:

1. $\sqrt{x^2 + 2x} = \sqrt{x^2(1 + \frac{2}{x})} = x\sqrt{1 + \frac{2}{x}} \approx x\left(1 + \frac{1}{x}\right) = x + 1$, since for large $x$, $\sqrt{1 + \frac{2}{x}} \approx 1 + \frac{1}{x}$.
2. $\sqrt{x^2 + x + x} = \sqrt{x^2 + 2x} \approx x + 1$, using the same approximation as in (1).

Step 2: Substitute the approximations

Now substitute these approximations into the limit expression: $\lim_{x \to +\infty} x \left( (x + 1) - 2(x + 1) \right)$

Simplifying inside the parentheses: $(x + 1) - 2(x + 1) = x + 1 - 2x - 2 = -x - 1$

Thus, the expression becomes: $\lim_{x \to +\infty} x(-x - 1) = \lim_{x \to +\infty} (-x^2 - x)$

Step 3: Evaluate the limit

As $x \to +\infty$, $-x^2 - x$ clearly tends to $-\infty$.

Thus, the final result is:

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

Would you like further details or clarifications?

Related Questions:

1. How do we handle limits that approach infinity with multiple terms?
2. Why is the approximation of square roots valid for large values of $x$?
3. How would the answer change if the coefficients inside the square roots were different?
4. What if the function inside the square roots had higher powers of $x$?
5. How does this technique apply to functions involving cube roots or other powers?

Tip:

When evaluating limits at infinity, approximate the dominant terms (highest powers) to simplify the expression and focus on leading behaviors.