To solve this problem, let's carefully analyze the given function and the corresponding limit.

Given:

The function $f(x) = \sqrt{x} + 8$.

We are asked to evaluate the limit:

$\lim_{h \to 0} \frac{f(6 + h) - f(6)}{h}$

Step-by-Step Solution:

1. Find $f(6)$:

   $f(6) = \sqrt{6} + 8$

2. Find $f(6 + h)$:

   $f(6 + h) = \sqrt{6 + h} + 8$

3. Set up the difference quotient:

   $\frac{f(6 + h) - f(6)}{h} = \frac{(\sqrt{6 + h} + 8) - (\sqrt{6} + 8)}{h}$

   Simplify the expression:

   $= \frac{\sqrt{6 + h} - \sqrt{6}}{h}$

4. Rationalize the numerator:

   Multiply the numerator and the denominator by the conjugate of the numerator:

   $\frac{\sqrt{6 + h} - \sqrt{6}}{h} \cdot \frac{\sqrt{6 + h} + \sqrt{6}}{\sqrt{6 + h} + \sqrt{6}}$

   This simplifies to:

   $= \frac{(6 + h) - 6}{h(\sqrt{6 + h} + \sqrt{6})}$

   $= \frac{h}{h(\sqrt{6 + h} + \sqrt{6})}$

   Simplify further:

   $= \frac{1}{\sqrt{6 + h} + \sqrt{6}}$

5. Evaluate the limit as $h \to 0$:

   Substitute $h = 0$:

   $\lim_{h \to 0} \frac{1}{\sqrt{6 + h} + \sqrt{6}} = \frac{1}{\sqrt{6} + \sqrt{6}} = \frac{1}{2\sqrt{6}}$

6. Simplify the final result:

   Since the answer is in its simplest form, we find:

   $\frac{1}{2\sqrt{6}}$

   However, the entered answer $\frac{1}{4}$ was incorrect. The correct answer is:

   $\frac{1}{2\sqrt{6}}$

Would you like further details, or do you have any questions about this explanation?

Related Questions to Explore:

1. How can we use the definition of a derivative to interpret this limit?
2. What happens if the function is different, such as $f(x) = x^2 + 8$?
3. How does rationalizing the numerator help in finding limits?
4. What are some common mistakes to avoid when simplifying difference quotients?
5. How can this problem be solved using alternative methods, like L'Hopital's Rule?

Tip:

When dealing with limits involving square roots, rationalizing the numerator is often a helpful technique to simplify the expression.