The given expression asks for the difference quotient, which is a form of the derivative definition, for the function $f(x) = \frac{1}{x}$ at $x = 2$.

We are asked to evaluate:

$\frac{f(2+h) - f(2)}{h}$

Step-by-step solution:

1. Substitute $f(x) = \frac{1}{x}$ into the expression: $f(2+h) = \frac{1}{2+h}, \quad f(2) = \frac{1}{2}$

2. Rewrite the difference quotient: $\frac{f(2+h) - f(2)}{h} = \frac{\frac{1}{2+h} - \frac{1}{2}}{h}$

3. Simplify the numerator: To combine the two fractions, we need a common denominator: $\frac{1}{2+h} - \frac{1}{2} = \frac{2 - (2+h)}{(2+h)(2)} = \frac{-h}{2(2+h)}$

4. Substitute this back into the difference quotient: $\frac{\frac{-h}{2(2+h)}}{h} = \frac{-h}{h \cdot 2(2+h)} = \frac{-1}{2(2+h)}$

5. As $h \to 0$, the expression becomes: $\frac{-1}{2(2+0)} = \frac{-1}{4}$

Thus, the value of the difference quotient is $\frac{-1}{4}$.

Would you like further details or clarifications on any step?

Here are some related questions you might find useful:

1. What is the general form of the difference quotient?
2. How does the difference quotient relate to the derivative?
3. What is the derivative of $f(x) = \frac{1}{x}$?
4. Why does the limit as $h \to 0$ give the derivative?
5. How can this process be applied to other functions?

Tip: The difference quotient is foundational to understanding how derivatives work, as it gives an approximation of the rate of change over small intervals.