The given limit problem is:

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

Step-by-Step Solution:

1. Expand the term $(h + 4)^2$: $(h + 4)^2 = h^2 + 8h + 16$

2. Substitute this into the limit expression: $\lim_{h \to 0} \frac{(h^2 + 8h + 16) - 16}{h}$

3. Simplify the numerator: $(h^2 + 8h + 16) - 16 = h^2 + 8h$ So the limit becomes: $\lim_{h \to 0} \frac{h^2 + 8h}{h}$

4. Factor out $h$ from the numerator: $\frac{h^2 + 8h}{h} = h + 8$

5. Now evaluate the limit: $\lim_{h \to 0} (h + 8) = 8$

Final Answer:

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

Let me know if you'd like further details or have any questions!

Here are 5 related questions to expand this concept:

1. How would you evaluate the limit $\lim_{h \to 0} \frac{(h + a)^2 - a^2}{h}$ for any constant $a$?
2. Can you apply the same limit technique to $\lim_{h \to 0} \frac{(h + 3)^3 - 27}{h}$?
3. How do you determine if a function is continuous at a point using limits?
4. What is the significance of limits in the definition of the derivative?
5. How does the binomial expansion help in solving limits involving powers?

Tip: When solving limits, always check if direct substitution gives an indeterminate form (like $\frac{0}{0}$) before applying algebraic techniques like factoring or expanding.