Let's evaluate the given limit expression step by step. The expression is:

$\lim_{{x \to 0}} \frac{x}{{\frac{1}{{2 + x}} - \frac{1}{2}}}$

Step 1: Simplify the denominator

The denominator is a difference of two fractions, $\frac{1}{{2 + x}} - \frac{1}{2}$. To subtract these fractions, we need to find a common denominator:

$\frac{1}{{2 + x}} - \frac{1}{2} = \frac{2 - (2 + x)}{2(2 + x)} = \frac{2 - 2 - x}{2(2 + x)} = \frac{-x}{2(2 + x)}$

So, the expression becomes:

$\lim_{{x \to 0}} \frac{x}{\frac{-x}{2(2 + x)}}$

Step 2: Simplify the fraction

Now, simplify the complex fraction by multiplying $x$ by the reciprocal of $\frac{-x}{2(2 + x)}$:

$\frac{x}{\frac{-x}{2(2 + x)}} = x \cdot \frac{2(2 + x)}{-x} = \frac{2(2 + x)}{-1} = -2(2 + x)$

Step 3: Evaluate the limit

Now, take the limit as $x$ approaches 0:

$\lim_{{x \to 0}} -2(2 + x) = -2(2 + 0) = -4$

Thus, the correct value of the limit is:

$\boxed{-4}$

The correct answer is $-4$, not $\frac{1}{4}$.

Would you like any further explanation or more details?

Here are 5 related questions for further practice:

1. What is the limit of $\lim_{{x \to 0}} \frac{x}{{x + 1}}$?
2. How would you evaluate $\lim_{{x \to 0}} \frac{x}{{(x^2 + 1) - 1}}$?
3. Can you solve $\lim_{{x \to 0}} \frac{\sin(x)}{x}$?
4. How do you handle indeterminate forms when evaluating limits?
5. What is the limit of $\lim_{{x \to 1}} \frac{1 - \cos(x)}{x - 1}$?

Tip: Always simplify the expression as much as possible before applying the limit, and check for indeterminate forms like $\frac{0}{0}$ where you might need additional techniques such as factoring or L'Hopital's Rule.