We are asked to evaluate the limit: $\lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1+\cos x}}{\sin^2 x}$

1. Problem Description

We are asked to evaluate the limit:

$\lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1+\cos x}}{\sin^2 x}$

2. Solution Steps

First, we multiply the numerator and denominator by $\sqrt{2} + \sqrt{1+\cos x}$:

$\lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1+\cos x}}{\sin^2 x} = \lim_{x \to 0} \frac{(\sqrt{2} - \sqrt{1+\cos x})(\sqrt{2} + \sqrt{1+\cos x})}{\sin^2 x (\sqrt{2} + \sqrt{1+\cos x})} = \lim_{x \to 0} \frac{2 - (1+\cos x)}{\sin^2 x (\sqrt{2} + \sqrt{1+\cos x})}$

$= \lim_{x \to 0} \frac{1 - \cos x}{\sin^2 x (\sqrt{2} + \sqrt{1+\cos x})}$

We can use the identity $\sin^2 x = 1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$:

$\lim_{x \to 0} \frac{1 - \cos x}{(1 - \cos x)(1 + \cos x) (\sqrt{2} + \sqrt{1+\cos x})} = \lim_{x \to 0} \frac{1}{(1 + \cos x)(\sqrt{2} + \sqrt{1+\cos x})}$

Now, as $x \to 0$, $\cos x \to 1$, so

$\lim_{x \to 0} \frac{1}{(1 + \cos x)(\sqrt{2} + \sqrt{1+\cos x})} = \frac{1}{(1 + 1)(\sqrt{2} + \sqrt{1+1})} = \frac{1}{2(\sqrt{2} + \sqrt{2})} = \frac{1}{2(2\sqrt{2})} = \frac{1}{4\sqrt{2}}$

We can rationalize the denominator:

$\frac{1}{4\sqrt{2}} = \frac{\sqrt{2}}{4\sqrt{2}\sqrt{2}} = \frac{\sqrt{2}}{4(2)} = \frac{\sqrt{2}}{8}$

3. Final Answer

$\frac{\sqrt{2}}{8}$