To solve the limit problem, we need to evaluate:

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

Step 1: Expand $(h + 4)^2$

Expanding $(h + 4)^2$ gives:

$(h + 4)^2 = h^2 + 8h + 16$

Step 2: Substitute and Simplify

Now, substitute this expansion back into the expression:

$\frac{(h + 4)^2 - 16}{h} = \frac{h^2 + 8h + 16 - 16}{h} = \frac{h^2 + 8h}{h}$

Step 3: Factor and Cancel

Now, factor $h$ out of the numerator:

$\frac{h(h + 8)}{h}$

Since $h \neq 0$ in the limit process, we can cancel $h$ from the numerator and the denominator:

$= h + 8$

Step 4: Apply the Limit

Now, take the limit as $h$ approaches 0:

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

Answer

The answer is $\boxed{8}$, which corresponds to option (e).

Would you like a detailed explanation of any particular step?

Here are some related questions for further exploration:

1. How do you expand and simplify expressions involving limits?
2. Why can we cancel $h$ in the fraction $\frac{h(h + 8)}{h}$?
3. How do we handle cases where direct substitution in limits yields an indeterminate form?
4. What is the purpose of factoring in limit problems like these?
5. How do limits apply to derivatives in calculus?

Tip: When you encounter indeterminate forms in limits, try factoring or using algebraic simplification to resolve them.