To evaluate the limit

$\lim_{h \to 0} \frac{1}{h} \left( \frac{1}{5+h} - \frac{1}{5} \right),$

let's go through the steps in detail:

1. Simplify the expression inside the parentheses:

   $\frac{1}{5+h} - \frac{1}{5}.$

   To combine these fractions, we find a common denominator, which is $5(5 + h)$:

   $\frac{1}{5+h} - \frac{1}{5} = \frac{5 - (5 + h)}{5(5 + h)} = \frac{5 - 5 - h}{5(5 + h)} = \frac{-h}{5(5 + h)}.$

2. Substitute this into the limit expression:

   $\lim_{h \to 0} \frac{1}{h} \cdot \frac{-h}{5(5 + h)}.$

3. Simplify by canceling $h$ from the numerator and denominator:

   $= \lim_{h \to 0} \frac{-h}{h \cdot 5(5 + h)} = \lim_{h \to 0} \frac{-1}{5(5 + h)}.$

4. Take the limit as $h \to 0$:

   As $h$ approaches 0, $5 + h$ approaches 5, so

   $= \frac{-1}{5 \cdot 5} = \frac{-1}{25}.$

Therefore,

$\lim_{h \to 0} \frac{1}{h} \left( \frac{1}{5+h} - \frac{1}{5} \right) = -\frac{1}{25}.$

Would you like a deeper explanation of any part of this solution?

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Here are some follow-up questions to deepen your understanding:

1. How does the choice of a common denominator simplify the problem?
2. Why does $h$ cancel out in the fraction?
3. What would happen if we changed the limit to $h \to \infty$?
4. How would the answer change if the denominator were $6 + h$ instead of $5 + h$?
5. Can you apply this technique to find limits of other similar rational functions?

Tip: Simplifying expressions with common denominators is a powerful technique for evaluating limits involving fractions.