The given problem is to solve the limit:

$\lim_{{x \to \frac{1}{2}}} \frac{4x^2 - 1}{x - \frac{1}{2}}.$

Step-by-Step Solution:

1. Substitution Check: First, let's directly substitute $x = \frac{1}{2}$ into the expression:

   $4\left(\frac{1}{2}\right)^2 - 1 = 4 \cdot \frac{1}{4} - 1 = 1 - 1 = 0.$

   The denominator: $x - \frac{1}{2} = \frac{1}{2} - \frac{1}{2} = 0.$

   This gives a $\frac{0}{0}$ indeterminate form, so we need to simplify the expression further.

2. Factorization: The numerator $4x^2 - 1$ can be factored as a difference of squares: $4x^2 - 1 = (2x - 1)(2x + 1).$

   The limit now becomes: $\lim_{{x \to \frac{1}{2}}} \frac{(2x - 1)(2x + 1)}{x - \frac{1}{2}}.$

3. Cancelation: Notice that $x - \frac{1}{2}$ is the same as $2x - 1$, so we can cancel the common term from the numerator and denominator: $\lim_{{x \to \frac{1}{2}}} \frac{(2x - 1)(2x + 1)}{2x - 1} = \lim_{{x \to \frac{1}{2}}} (2x + 1).$

4. Final Substitution: Now substitute $x = \frac{1}{2}$ into the simplified expression: $2\left(\frac{1}{2}\right) + 1 = 1 + 1 = 2.$

Thus, the value of the limit is:

$\boxed{2}.$

Would you like further details or have any questions?

Here are some related questions:

1. What is L'Hopital's Rule, and how can it help in solving limits with indeterminate forms?
2. How do you handle limits that result in a $\frac{0}{0}$ form?
3. Can every indeterminate form be simplified through algebraic techniques like factoring?
4. What is the geometric interpretation of a limit?
5. How does the concept of continuity relate to limits?

Tip: Always check for factorizations, especially in indeterminate forms, as this often simplifies the limit evaluation.