We are tasked with finding the following limit from the image you provided:

$\lim_{{x \to 0.5^-}} \frac{{2x - 1}}{{2x^3 - x^2}}$

Step-by-step Solution:

1. Substitute $x = 0.5$ directly into the expression to check if it results in an indeterminate form:

   $\text{Numerator: } 2(0.5) - 1 = 1 - 1 = 0$ $\text{Denominator: } 2(0.5)^3 - (0.5)^2 = 2(0.125) - 0.25 = 0.25 - 0.25 = 0$

   Since both the numerator and denominator result in 0, we have an indeterminate form $\frac{0}{0}$. Thus, we need to simplify the expression further by factoring or using another method.

2. Factor the expression:

   Let's try to factor the denominator $2x^3 - x^2$. First, we can factor out $x^2$:

   $2x^3 - x^2 = x^2(2x - 1)$

3. Rewrite the original limit expression:

   Now we rewrite the limit with the factored denominator:

   $\lim_{{x \to 0.5^-}} \frac{{2x - 1}}{{x^2(2x - 1)}}$

4. Cancel out common factors:

   Notice that $2x - 1$ is common in both the numerator and the denominator, so we can cancel it out:

   $\lim_{{x \to 0.5^-}} \frac{1}{x^2}$

5. Evaluate the simplified limit:

   Now, evaluate $\lim_{{x \to 0.5^-}} \frac{1}{x^2}$:

   $\frac{1}{(0.5)^2} = \frac{1}{0.25} = 4$

Thus, the value of the limit is:

$\boxed{4}$

Would you like more details or explanations on any part of this solution?

Here are 5 related questions to expand your understanding:

1. How would you approach the limit if $x \to 0.5^+$ (from the right)?
2. What are common techniques to deal with indeterminate forms in limits?
3. How would you find the limit if the denominator did not factor as easily?
4. Can you apply L'Hôpital's Rule to this problem instead of factoring?
5. What is the significance of factoring in simplifying limits?

Tip: When you encounter an indeterminate form $\frac{0}{0}$, it's a sign to apply techniques like factoring, L'Hôpital's Rule, or algebraic simplification to resolve the limit.