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To solve the limit

\[
\lim_{{x \to -1}} \frac{\sqrt{2x + 3} - \sqrt{2x^2 - 1}}{x + 1},
\]

we notice that directly substituting \( x = -1 \) results in an indeterminate form \(\frac{0}{0}\). To resolve this, let's use the technique of rationalizing the numerator by multiplying both the numerator and the denominator by the conjugate of the numerator, which is \(\sqrt{2x + 3} + \sqrt{2x^2 - 1}\).

### Step-by-Step Solution:

1. **Multiply by the Conjugate:**

   Multiply the numerator and the denominator by \(\sqrt{2x + 3} + \sqrt{2x^2 - 1}\):

   \[
   \lim_{{x \to -1}} \frac{(\sqrt{2x + 3} - \sqrt{2x^2 - 1})(\sqrt{2x + 3} + \sqrt{2x^2 - 1})}{(x + 1)(\sqrt{2x + 3} + \sqrt{2x^2 - 1})}.
   \]

2. **Simplify the Numerator:**

   Using the difference of squares formula \((a - b)(a + b) = a^2 - b^2\), we get:

   \[
   = \lim_{{x \to -1}} \frac{(2x + 3) - (2x^2 - 1)}{(x + 1)(\sqrt{2x + 3} + \sqrt{2x^2 - 1})}.
   \]

3. **Expand and Simplify:**

   Simplify the expression in the numerator:

   \[
   = \lim_{{x \to -1}} \frac{2x + 3 - 2x^2 + 1}{(x + 1)(\sqrt{2x + 3} + \sqrt{2x^2 - 1})}.
   \]

   Combine like terms:

   \[
   = \lim_{{x \to -1}} \frac{-2x^2 + 2x + 4}{(x + 1)(\sqrt{2x + 3} + \sqrt{2x^2 - 1})}.
   \]

4. **Factor the Numerator:**

   Factor \(-2x^2 + 2x + 4\):

   \[
   = \lim_{{x \to -1}} \frac{-2(x + 1)(x - 2)}{(x + 1)(\sqrt{2x + 3} + \sqrt{2x^2 - 1})}.
   \]

   Cancel out the \((x + 1)\) term from the numerator and denominator:

   \[
   = \lim_{{x \to -1}} \frac{-2(x - 2)}{\sqrt{2x + 3} + \sqrt{2x^2 - 1}}.
   \]

5. **Substitute \( x = -1 \):**

   Now, we can substitute \( x = -1 \):

   \[
   = \frac{-2(-1 - 2)}{\sqrt{2(-1) + 3} + \sqrt{2(-1)^2 - 1}}.
   \]

6. **Calculate the Terms:**

   Simplify each part:

   - For the numerator: \( -2(-3) = 6 \).
   - For the first square root: \(\sqrt{2(-1) + 3} = \sqrt{-2 + 3} = \sqrt{1} = 1\).
   - For the second square root: \(\sqrt{2(-1)^2 - 1} = \sqrt{2 \cdot 1 - 1} = \sqrt{1} = 1\).

   So we get:

   \[
   = \frac{6}{1 + 1} = \frac{6}{2} = 3.
   \]

### Final Answer:

\[
\lim_{{x \to -1}} \frac{\sqrt{2x + 3} - \sqrt{2x^2 - 1}}{x + 1} = 3.
\]

Would you like more details, or do you have any questions?

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Here are 5 related questions you might find helpful:

1. What are the conditions for using the conjugate method in limits?
2. How does one identify when to use rationalization versus other limit-solving techniques?
3. What is the significance of the indeterminate form \(\frac{0}{0}\) in limits?
4. How does factoring help simplify limit expressions?
5. Can we apply L'Hôpital's Rule to solve this limit, and how would that work?

**Tip:** When you encounter square roots in limits, rationalizing the numerator or denominator can often simplify the expression and help resolve indeterminate forms.

Evaluate the limit \( \lim_{{x \to -1}} \frac{\sqrt{2x + 3} - \sqrt{2x^2 - 1}}{x + 1} \).

Learn how to evaluate the limit \( \lim_{{x \to -1}} \frac{\sqrt{2x + 3} - \sqrt{2x^2 - 1}}{x + 1} \) by rationalizing the numerator. This problem covers key concepts in limits, indeterminate forms, and the difference of squares formula. Suitable for advanced high school students in Grades 11-12.

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