1. **State the problem:** We want to evaluate the limit $$\lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}.$$\n\n2. **Recall the formula:** This limit resembles the definition of the derivative of the function $f(x) = \frac{1}{x}$ at the point $x$. The derivative is given by $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.$$\n\n3. **Simplify the expression inside the limit:**\n\n$$\frac{\frac{1}{x+h} - \frac{1}{x}}{h} = \frac{\frac{x - (x+h)}{x(x+h)}}{h} = \frac{\frac{-h}{x(x+h)}}{h}.$$\n\n4. **Simplify the complex fraction:**\n\n$$= \frac{-h}{x(x+h)} \cdot \frac{1}{h} = \frac{-1}{x(x+h)}.$$\n\n5. **Evaluate the limit as $h \to 0$:**\n\n$$\lim_{h \to 0} \frac{-1}{x(x+h)} = \frac{-1}{x \cdot x} = \frac{-1}{x^2}.$$\n\n6. **Interpretation:** The limit equals the derivative of $f(x) = \frac{1}{x}$, which is $f'(x) = -\frac{1}{x^2}$.\n\n**Final answer:** $$\lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h} = -\frac{1}{x^2}.$$