We are tasked with evaluating the following limit expression:

$\lim_{h \to 0} \frac{(2 + h)^{-1} - 2^{-1}}{h}$

Step-by-Step Solution:

1. Rewrite the expression: $\lim_{h \to 0} \frac{\frac{1}{2 + h} - \frac{1}{2}}{h}$

2. Find a common denominator for the terms inside the numerator: The common denominator between $2 + h$ and $2$ is $2(2 + h)$. So: $\frac{1}{2 + h} - \frac{1}{2} = \frac{2 - (2 + h)}{2(2 + h)} = \frac{2 - 2 - h}{2(2 + h)} = \frac{-h}{2(2 + h)}$

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

4. Simplify the expression: $\lim_{h \to 0} \frac{-h}{h \cdot 2(2 + h)}$

   The $h$ in the numerator and denominator cancels out: $\lim_{h \to 0} \frac{-1}{2(2 + h)}$

5. Evaluate the limit as $h \to 0$: As $h$ approaches 0, $2 + h$ approaches 2. Therefore: $\frac{-1}{2(2 + 0)} = \frac{-1}{4}$

Final Answer:

$\lim_{h \to 0} \frac{(2 + h)^{-1} - 2^{-1}}{h} = \frac{-1}{4}$

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Would you like more details on any of these steps? Here are five related questions to explore:

1. How does the limit definition relate to the derivative of a function?
2. What does the result $\frac{-1}{4}$ represent geometrically for this specific function?
3. How do you handle limits involving more complex algebraic fractions?
4. What happens if we change the constants in this type of limit problem?
5. How do you apply L'Hopital's Rule to similar limit problems?

Tip: Always simplify the algebraic expression before evaluating the limit to avoid unnecessary complexity.