https://file.aiquickdraw.com/mathbot/file/17264565752112skfe1q0.jpg

Let's go through the steps shown in the image to evaluate the limit:

We are asked to evaluate the limit:

\[
\lim_{h \to 0} \frac{(4 + h)^2 - 16}{h}
\]

### Step 1: Define \( F(h) \)
Define \( F(h) \) as the expression inside the limit:

\[
F(h) = \frac{(4 + h)^2 - 16}{h}
\]

### Step 2: Expand \( (4 + h)^2 \)
Next, expand \( (4 + h)^2 \):

\[
(4 + h)^2 = 16 + 8h + h^2
\]

Now substitute this into \( F(h) \):

\[
F(h) = \frac{16 + 8h + h^2 - 16}{h}
\]

Simplify the numerator:

\[
F(h) = \frac{8h + h^2}{h}
\]

### Step 3: Simplify the Expression
Factor the numerator:

\[
F(h) = \frac{h(8 + h)}{h}
\]

For \( h \neq 0 \), we can cancel \( h \) from the numerator and denominator:

\[
F(h) = 8 + h
\]

### Step 4: Take the Limit as \( h \to 0 \)
Now, take the limit of \( F(h) \) as \( h \to 0 \):

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

### Final Answer:
The limit is:

\[
\lim_{h \to 0} \frac{(4 + h)^2 - 16}{h} = 8
\]

---

Would you like more details about any step or have further questions?

Here are 5 related questions for deeper understanding:
1. How would you approach this problem using L'Hôpital's Rule?
2. Why can't we directly substitute \( h = 0 \) at the start?
3. What does the cancellation of \( h \) in the numerator and denominator represent geometrically?
4. How does this limit relate to the concept of derivatives?
5. How can we apply similar techniques to evaluate the limit of a different quadratic expression?

**Tip:** When simplifying limits involving a difference of squares, expanding the binomial and factoring terms is often key to canceling problematic terms.

Evaluate the limit: lim_{h \to 0} \frac{(4 + h)^2 - 16}{h}

This problem evaluates the limit of the expression (4 + h)^2 - 16 over h as h approaches 0, using quadratic expansion and algebraic simplification techniques. Suitable for students in advanced algebra or early calculus.

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