This mathematical problem involves evaluating the limit:

$\lim_{x \to 0} \frac{\sin x}{\sqrt{1 - \cos x}}.$

Let me guide you through the solution.

Step 1: Simplify the denominator

Recall the trigonometric identity: $1 - \cos x = 2\sin^2\left(\frac{x}{2}\right).$ Substituting this into the denominator, we have: $\sqrt{1 - \cos x} = \sqrt{2\sin^2\left(\frac{x}{2}\right)} = \sqrt{2} \cdot \left|\sin\left(\frac{x}{2}\right)\right|.$

Step 2: Rewrite the limit

The limit now becomes: $\lim_{x \to 0} \frac{\sin x}{\sqrt{2} \cdot \left|\sin\left(\frac{x}{2}\right)\right|}.$

Step 3: Handle the absolute value

For $x \to 0$, $\sin\left(\frac{x}{2}\right) \to 0$ and is positive. Thus, $\left|\sin\left(\frac{x}{2}\right)\right| = \sin\left(\frac{x}{2}\right)$. The limit simplifies to: [ \lim_{x \to 0} \frac{\sin x}{\sqrt{2} \cdot \sin\left(\frac{x}{2