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The given problem is to find the limit:

\[
\lim_{{h \to 0}} \frac{\sqrt{25 + h} - 5}{h}
\]

### Step-by-Step Solution:

1. **Substitute** \( h = 0 \) directly to see if we get an indeterminate form:
   \[
   \frac{\sqrt{25 + 0} - 5}{0} = \frac{\sqrt{25} - 5}{0} = \frac{5 - 5}{0} = \frac{0}{0}
   \]
   This is an indeterminate form \( \frac{0}{0} \), so we must simplify the expression.

2. **Rationalizing the numerator**: Multiply both the numerator and denominator by the conjugate of the numerator, which is \( \sqrt{25 + h} + 5 \):
   \[
   \frac{\sqrt{25 + h} - 5}{h} \cdot \frac{\sqrt{25 + h} + 5}{\sqrt{25 + h} + 5} = \frac{(\sqrt{25 + h})^2 - 5^2}{h(\sqrt{25 + h} + 5)}
   \]
   This simplifies the numerator using the difference of squares:
   \[
   = \frac{25 + h - 25}{h(\sqrt{25 + h} + 5)} = \frac{h}{h(\sqrt{25 + h} + 5)}
   \]

3. **Cancel the \( h \) terms**:
   \[
   = \frac{1}{\sqrt{25 + h} + 5}
   \]

4. **Take the limit as \( h \to 0 \)**:
   \[
   \lim_{{h \to 0}} \frac{1}{\sqrt{25 + h} + 5} = \frac{1}{\sqrt{25} + 5} = \frac{1}{5 + 5} = \frac{1}{10}
   \]

Thus, the solution is:

\[
\lim_{{h \to 0}} \frac{\sqrt{25 + h} - 5}{h} = \frac{1}{10}
\]

Would you like further explanation or any additional questions? Here are some related questions to deepen understanding:

1. What happens if you don't rationalize the numerator in a limit problem?
2. How can you apply this method to similar limit problems involving radicals?
3. What is the difference of squares formula, and why is it useful in this context?
4. How does this problem relate to the definition of a derivative?
5. What other methods can be used to evaluate limits that result in indeterminate forms?

**Tip**: Always check if the direct substitution gives an indeterminate form before applying algebraic techniques to simplify.

Find the limit lim(h→0) ((√(25 + h) - 5) / h).

Solve the limit lim(h→0) ((√(25 + h) - 5) / h) by rationalizing the numerator and simplifying the expression. This example covers algebraic techniques like the difference of squares and rationalization. Suitable for students learning calculus.

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