$$ \lim_{h\rarr0} $$$$ \frac{1}{h} $$($$ \frac{1}{5+h}-\frac15 $$)

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?

---

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.

Evaluate the limit $ \lim_{h \to 0} \frac{1}{h} \left( \frac{1}{5+h} - \frac{1}{5} \right) $.

Explore the solution to $ \lim_{h \to 0} \frac{1}{h} \left( \frac{1}{5+h} - \frac{1}{5} \right) $ with step-by-step instructions. Learn about calculus concepts like limits, algebraic simplification, and limit laws. Ideal for students in advanced high school or introductory college calculus.

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