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The given expression:

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

resembles the definition of the derivative:

\[
f'(a) = \lim_{{h \to 0}} \frac{f(a + h) - f(a)}{h}
\]

In this case, we can identify \( f(x) \) and \( a \) by comparing the structure of the limit expression.

1. **Finding \( f(x) \):**
   Notice that the term \((4 + h)^2\) suggests that \( f(x) = x^2 \) because the expression \((4 + h)^2\) matches the form \( f(4 + h) \) for \( f(x) = x^2 \).

2. **Finding \( a \):**
   Since the expression is evaluated at \( 4 + h \), it indicates that \( a = 4 \).

So:
- \( f(x) = x^2 \)
- \( a = 4 \)

This setup allows us to express the limit as the derivative of \( f(x) = x^2 \) at \( x = 4 \).

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

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**Related Questions:**
1. How would you evaluate the limit if \( f(x) = x^3 \) and \( a = 4 \)?
2. What would change if the function \( f(x) \) was \( \sqrt{x} \) and \( a = 4 \)?
3. How does this limit expression change if we evaluate \( f(x) = x^2 \) at \( a = 5 \)?
4. What is the purpose of the derivative definition in calculus?
5. How can you confirm \( f(x) \) and \( a \) from similar derivative limit expressions?

**Tip:** To solve these types of problems, always compare the expression with the general derivative formula \( f'(a) = \lim_{{h \to 0}} \frac{f(a + h) - f(a)}{h} \) to identify the function \( f(x) \) and point \( a \).

The limit \( \lim_{h \to 0} \frac{(4 + h)^2 - 16}{h} \) is the standard expression defining the derivative of some function \( f(x) \) at some number \( a \). Find \( f \) and \( a \).

Solve the limit \( \lim_{h \to 0} \frac{(4 + h)^2 - 16}{h} \) to identify the function \( f(x) \) and the point \( a \) in the context of derivatives. Learn the application of derivative definition and limits with step-by-step guidance. Suitable for students in Grades 11-12.

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