We need to evaluate the limit of the following expression as $s$ approaches $-1$:

$\lim_{s \to -1} \frac{2 - \sqrt{s^2 + 3}}{1 - s^2}$

Step 1: Substitute $s = -1$

Substitute $s = -1$ directly into the expression:

$\frac{2 - \sqrt{(-1)^2 + 3}}{1 - (-1)^2} = \frac{2 - \sqrt{1 + 3}}{1 - 1} = \frac{2 - \sqrt{4}}{0} = \frac{2 - 2}{0} = \frac{0}{0}$

This gives us an indeterminate form $\frac{0}{0}$, so we can apply L'Hôpital's Rule to find the limit.

Step 2: Apply L'Hôpital's Rule

L'Hôpital's Rule states that if a limit results in the indeterminate form $\frac{0}{0}$, we can differentiate the numerator and the denominator separately and then evaluate the limit.

Differentiate the numerator:

$\frac{d}{ds} \left( 2 - \sqrt{s^2 + 3} \right) = 0 - \frac{1}{2\sqrt{s^2 + 3}} \cdot 2s = -\frac{s}{\sqrt{s^2 + 3}}$

Differentiate the denominator:

$\frac{d}{ds} \left( 1 - s^2 \right) = -2s$

Now, the limit becomes: $\lim_{s \to -1} \frac{-\frac{s}{\sqrt{s^2 + 3}}}{-2s} = \lim_{s \to -1} \frac{\frac{s}{\sqrt{s^2 + 3}}}{2s}$

Step 3: Simplify and Evaluate the Limit

Cancel the $s$ terms (note $s \neq 0$ in this case):

$\lim_{s \to -1} \frac{1}{2\sqrt{s^2 + 3}} = \frac{1}{2\sqrt{(-1)^2 + 3}} = \frac{1}{2\sqrt{4}} = \frac{1}{2 \times 2} = \frac{1}{4}$

Thus, the limit is:

$\boxed{\frac{1}{4}}$

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Would you like more details or have any further questions?

Related questions:

1. How does L'Hôpital's Rule work for other indeterminate forms like $\frac{\infty}{\infty}$?
2. What are other methods to evaluate limits apart from L'Hôpital's Rule?
3. Can L'Hôpital’s Rule be applied if both numerator and denominator are not differentiable?
4. What is the significance of the indeterminate form $\frac{0}{0}$?
5. How can we simplify limits involving square roots without using L'Hôpital's Rule?

Tip: When using L'Hôpital’s Rule, always verify that you have an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before differentiating.