To find the limit of the function \(\frac{\sqrt{4 - x} - 2}{x}\) as \(x\) approaches 0, we first recognize that direct substitution leads to an indeterminate form \(\frac{0}{0}\). In such cases, using the conjugate can simplify the expression. The conjugate of the numerator, \(\sqrt{4 - x} - 2\), is \(\sqrt{4 - x} + 2\).

We multiply both the numerator and the denominator by this conjugate:

\[\frac{(\sqrt{4 - x} - 2)(\sqrt{4 - x} + 2)}{x(\sqrt{4 - x} + 2)}\]

Expanding the numerator using the difference of squares gives:

\[(\sqrt{4 - x})^2 - 2^2 = (4 - x) - 4 = -x\]

Thus, the expression simplifies to:

\[\frac{-x}{x(\sqrt{4 - x} + 2)}\]

We can cancel \(x\) in the numerator and denominator (noting that \(x \neq 0\) in the limit process), leading to:

\[\frac{-1}{\sqrt{4 - x} + 2}\]

Now, we can safely substitute \(x = 0\) into the simplified expression:

\[\frac{-1}{\sqrt{4 - 0} + 2} = \frac{-1}{2 + 2} = \frac{-1}{4}\]

Therefore, the limit of the function as \(x\) approaches 0 is \(\frac{-1}{4}\).