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?

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.