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.