You took wrong question theri is square on √x + 2

Thank you for the clarification. Let's correct the interpretation of the problem based on your description.

The correct expression is:

\[
\lim_{x \to \infty} 2x \left(\sqrt{x + 1} - \sqrt{x + 2}\right)^2
\]

### Step 1: Simplify the Expression

Let's first examine the expression inside the limit:

\[
\sqrt{x + 1} - \sqrt{x + 2}
\]

As \( x \) becomes large, both \( \sqrt{x + 1} \) and \( \sqrt{x + 2} \) approach \( \sqrt{x} \), so the difference between them becomes small. To handle this, we'll again multiply and divide by the conjugate:

\[
\sqrt{x + 1} - \sqrt{x + 2} = \frac{(\sqrt{x + 1} - \sqrt{x + 2})(\sqrt{x + 1} + \sqrt{x + 2})}{\sqrt{x + 1} + \sqrt{x + 2}} = \frac{(x + 1) - (x + 2)}{\sqrt{x + 1} + \sqrt{x + 2}}
\]

Simplifying this gives:

\[
\sqrt{x + 1} - \sqrt{x + 2} = \frac{-1}{\sqrt{x + 1} + \sqrt{x + 2}}
\]

### Step 2: Square the Expression

Next, we square this difference:

\[
\left(\sqrt{x + 1} - \sqrt{x + 2}\right)^2 = \left(\frac{-1}{\sqrt{x + 1} + \sqrt{x + 2}}\right)^2 = \frac{1}{(\sqrt{x + 1} + \sqrt{x + 2})^2}
\]

### Step 3: Substitute and Simplify the Limit Expression

Now, substitute this back into the original limit expression:

\[
\lim_{x \to \infty} 2x \cdot \frac{1}{(\sqrt{x + 1} + \sqrt{x + 2})^2}
\]

Since \( \sqrt{x + 1} \) and \( \sqrt{x + 2} \) are approximately \( \sqrt{x} \) when \( x \) is large, we can approximate the sum:

\[
\sqrt{x + 1} + \sqrt{x + 2} \approx 2\sqrt{x}
\]

So the expression becomes:

\[
\lim_{x \to \infty} 2x \cdot \frac{1}{(2\sqrt{x})^2} = \lim_{x \to \infty} 2x \cdot \frac{1}{4x} = \lim_{x \to \infty} \frac{2x}{4x} = \lim_{x \to \infty} \frac{1}{2} = \frac{1}{2}
\]

### Conclusion

The correct limit is:

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

Would you like any further details or explanations on this?

Here are some related questions for further exploration:
1. How would the limit change if there was a cube instead of a square?
2. Can you calculate the limit as \( x \) approaches zero instead of infinity?
3. What happens if you change the coefficients in the limit expression?
4. How do you evaluate limits involving higher powers of \( x \)?
5. What is the significance of multiplying by the conjugate in limit problems?
6. How do limits behave in expressions with more than two terms under the square root?
7. Can L'Hopital's rule be used for this limit?
8. How does this approach help with evaluating indeterminate forms?

**Tip:** Always double-check the structure of the given mathematical expression before solving. Small changes like the placement of exponents can significantly affect the solution method.

Explore the solution to the limit problem involving the difference of square roots as x approaches infinity. Learn step-by-step how to simplify and calculate the limit using properties of square roots and limits. Suitable for advanced high school students and college-level learners.

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