What does lim h->  8(1/2 +h)^2  - 8(1/2)^2 / h

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.

What does lim h-> 8(1/2 + h)^2 - 8(1/2)^2 / h

Understand how to evaluate the limit lim h→0 (8(1/2 + h)^2 - 8(1/2)^2) / h using algebraic expansion and limit properties. This detailed solution walks through each step, making it perfect for high school or early college students learning calculus and limits.

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