To evaluate the limit:

$\lim_{{h \to 0}} \frac{8\left(\frac{1}{2} + h\right)^2 - 8\left(\frac{1}{2}\right)^2}{h}$

Step-by-step Solution:

1. Evaluate the two square terms:

   First, expand $\left(\frac{1}{2} + h\right)^2$:

   $\left(\frac{1}{2} + h\right)^2 = \left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right)h + h^2 = \frac{1}{4} + h + h^2$

   Then multiply by 8:

   $8\left(\frac{1}{2} + h\right)^2 = 8\left(\frac{1}{4} + h + h^2\right) = 2 + 8h + 8h^2$

2. Evaluate $8\left(\frac{1}{2}\right)^2$:

   $8\left(\frac{1}{2}\right)^2 = 8 \times \frac{1}{4} = 2$

3. Subtract the two terms:

   $8\left(\frac{1}{2} + h\right)^2 - 8\left(\frac{1}{2}\right)^2 = (2 + 8h + 8h^2) - 2 = 8h + 8h^2$

4. Simplify the limit expression:

   Now substitute into the original limit expression:

   $\lim_{{h \to 0}} \frac{8h + 8h^2}{h}$

   Factor the numerator:

   $\lim_{{h \to 0}} \frac{h(8 + 8h)}{h}$

   Cancel out the $h$ (as $h \neq 0$):

   $\lim_{{h \to 0}} (8 + 8h)$

5. Evaluate the limit as $h \to 0$:

   $\lim_{{h \to 0}} (8 + 8h) = 8$

Final Answer:

$\boxed{8}$

Would you like more details, or do you have any questions?

Here are 5 related questions to explore:

1. What happens if the coefficient inside the square is different (e.g., $\frac{1}{3}$ instead of $\frac{1}{2}$)?
2. How do limits behave if you add more terms involving $h^3$ or higher powers of $h$?
3. What is the geometric interpretation of the given limit?
4. Can you apply L'Hôpital's Rule to solve this limit?
5. How would the result change if the expression were inside a square root?

Tip: When evaluating limits with expressions that simplify to $\frac{0}{0}$, factoring or expanding is often useful to cancel terms.