To evaluate the given limit:

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

it is noted that the modulus on $|\sin x|$ can be simplified as $\sin x$ itself because $x$ approaches $0$ and $\sin x$ is positive in the vicinity of $x = 0$.

This reduces the expression to:

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

Now, let's simplify further. We use the trigonometric identity $1 - \cos x = 2 \sin^2 \frac{x}{2}$. Substituting this identity into the denominator:

$\sqrt{1 - \cos x} = \sqrt{2 \sin^2 \frac{x}{2}} = \sqrt{2} |\sin \frac{x}{2}|$

Since $x \to 0$, $\sin \frac{x}{2}$ is also positive. Thus, $|\sin \frac{x}{2}| = \sin \frac{x}{2}$. The limit becomes:

$\lim_{x \to 0} \frac{\sin x}{\sqrt{2} \sin \frac{x}{2}}$

Using the double-angle identity $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$, substitute for $\sin x$:

$\lim_{x \to 0} \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{\sqrt{2} \sin \frac{x}{2}}$

Cancel $\sin \frac{x}{2}$ in the numerator and denominator (it is non-zero for small $x$):

$\lim_{x \to 0} \frac{2 \cos \frac{x}{2}}{\sqrt{2}}$

As $x \to 0$, $\cos \frac{x}{2} \to \cos 0 = 1$. Thus, the limit evaluates to:

$\frac{2}{\sqrt{2}} = \sqrt{2}$

Final Answer:

$\lim_{x \to 0} \frac{|\sin x|}{\sqrt{1 - \cos x}} = \sqrt{2}$

Let me know if you have further questions, or if you want additional details on any step.

Here are some related questions for exploration:

1. What happens if the limit was approaching $-0$ instead of $0$?
2. How is the modulus affecting the behavior of $\sin x$ in different quadrants?
3. What other trigonometric identities might simplify similar expressions?
4. How can L’Hôpital’s Rule be applied to verify this result?
5. What happens if the denominator includes $1 + \cos x$ instead of $1 - \cos x$?

Tip: Always explore trigonometric identities when dealing with limits involving sine and cosine near $x = 0$.