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

We wish to evaluate the limit

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

First, to remove the square-root expression from the denominator we rationalise. The standard algebraic identity we use is

$$(a-b)(a+b)=a^{2}-b^{2}.$$

Here we take $$a=\sqrt{2} \quad\text{and}\quad b=\sqrt{1+\cos x}.$$ Multiplying numerator and denominator by $$a+b=\sqrt{2}+\sqrt{1+\cos x}$$ leaves the value of the fraction unchanged but simplifies the denominator.

So we write

$$\frac{\sin^{2}x}{\sqrt{2}-\sqrt{1+\cos x}} \;=\; \frac{\sin^{2}x\;(\sqrt{2}+\sqrt{1+\cos x})} {(\sqrt{2}-\sqrt{1+\cos x})(\sqrt{2}+\sqrt{1+\cos x})}.$$

Applying the identity just quoted to the denominator:

$$ (\sqrt{2})^{2}-(\sqrt{1+\cos x})^{2} \;=\; 2-(1+\cos x) \;=\; 1-\cos x.$$

Hence the limit becomes

$$\lim_{x\to0}\; \dfrac{\sin^{2}x\;(\sqrt{2}+\sqrt{1+\cos x})}{1-\cos x}.$$

Now we recall the Pythagorean identity

$$1-\cos x = 2\sin^{2}\left(\dfrac{x}{2}\right),$$

and also the double-angle relation

$$\sin x = 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right).$$

Squaring the double-angle formula gives

$$\sin^{2}x = 4\sin^{2}\left(\dfrac{x}{2}\right)\cos^{2}\left(\dfrac{x}{2}\right).$$

Substituting these two expressions into our fraction we get

$$\frac{\sin^{2}x}{1-\cos x} \;=\; \frac{4\sin^{2}\!\left(\dfrac{x}{2}\right)\cos^{2}\!\left(\dfrac{x}{2}\right)} {2\sin^{2}\!\left(\dfrac{x}{2}\right)} \;=\; 2\cos^{2}\!\left(\dfrac{x}{2}\right).$$

Therefore the entire expression simplifies to

$$\left[\,2\cos^{2}\!\left(\dfrac{x}{2}\right)\right] \; \bigl(\sqrt{2}+\sqrt{1+\cos x}\bigr).$$

Now we take the limit as $$x \to 0.$$ We know the standard limits

$$\cos\left(\dfrac{x}{2}\right) \xrightarrow[x\to0]{} 1,$$ $$\cos x \xrightarrow[x\to0]{} 1.$$

Using these values we have

$$2\cos^{2}\!\left(\dfrac{0}{2}\right) \;=\; 2\cdot 1^{2} \;=\; 2,$$

and

$$\sqrt{1+\cos 0} \;=\; \sqrt{1+1} \;=\; \sqrt{2}.$$

Thus the bracket evaluates to

$$\sqrt{2}+\sqrt{2}=2\sqrt{2}.$$

Multiplying the two limiting factors gives

$$2 \times 2\sqrt{2} \;=\; 4\sqrt{2}.$$

So the limit is

$$\boxed{4\sqrt{2}}.$$

Hence, the correct answer is Option A.