limit inf (x*sqrt(x+1)(1-sqrt(2x+3))/(7-6x+4x^2))

Let's find the limit as \( x \to \infty \) of the following expression:

\[
\lim_{x \to \infty} \frac{x \sqrt{x+1} (1 - \sqrt{2x+3})}{7 - 6x + 4x^2}
\]

### Step-by-step approach:

1. **Denominator Simplification:**
   - In the denominator \( 7 - 6x + 4x^2 \), the highest degree term is \( 4x^2 \). As \( x \to \infty \), this term will dominate, so we approximate the denominator as:
     \[
     4x^2.
     \]

2. **Numerator Simplification:**
   - In the numerator \( x \sqrt{x+1} (1 - \sqrt{2x+3}) \), let's analyze the two key parts:
     - \( \sqrt{x+1} \) can be approximated as \( \sqrt{x} \) for large \( x \), since:
       \[
       \sqrt{x+1} = \sqrt{x(1 + \frac{1}{x})} \approx \sqrt{x} \left(1 + \frac{1}{2x}\right) \approx \sqrt{x} \text{ as } x \to \infty.
       \]
     - \( 1 - \sqrt{2x+3} \): Using a binomial expansion for large \( x \), we get:
       \[
       \sqrt{2x+3} = \sqrt{2x}\left(1 + \frac{3}{2(2x)}\right) = \sqrt{2x} \left(1 + \frac{3}{4x}\right) \approx \sqrt{2x}.
       \]
       Thus, \( 1 - \sqrt{2x+3} \approx 1 - \sqrt{2x} \).

3. **Combine the Terms:**
   - Now, the numerator becomes:
     \[
     x \sqrt{x} (1 - \sqrt{2x}) = x^{3/2} (1 - \sqrt{2x}).
     \]
     As \( x \to \infty \), \( \sqrt{2x} \) grows much faster than 1, so \( 1 - \sqrt{2x} \) behaves like \( -\sqrt{2x} \). Hence, the numerator becomes approximately:
     \[
     -x^{3/2} \sqrt{2x} = -\sqrt{2} x^2.
     \]

4. **Final Simplification:**
   - The expression now looks like:
     \[
     \lim_{x \to \infty} \frac{-\sqrt{2} x^2}{4x^2}.
     \]
   - Simplifying:
     \[
     \frac{-\sqrt{2} x^2}{4x^2} = \frac{-\sqrt{2}}{4}.
     \]

### Final Answer:
\[
\lim_{x \to \infty} \frac{x \sqrt{x+1} (1 - \sqrt{2x+3})}{7 - 6x + 4x^2} = \frac{-\sqrt{2}}{4}.
\]

Would you like more details or have any questions?

Here are 5 related questions:
1. How do you apply L'Hopital's rule to limits involving infinity?
2. How do you simplify square roots involving large \( x \)?
3. What are some strategies for handling limits of rational functions at infinity?
4. How can you use binomial expansions for approximations?
5. Why do higher-degree terms dominate in limits as \( x \to \infty \)?

**Tip:** When simplifying limits, focus on the highest degree terms, as they often determine the behavior of the function at infinity.

Discover the step-by-step solution for finding the limit of the expression (x sqrt(x+1)(1 - sqrt(2x+3)))/(7 - 6x + 4x^2) as x approaches infinity. This advanced calculus problem involves simplification, binomial expansions, and rational function limits. Suitable for undergraduate or advanced high school calculus students.

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