We are asked to evaluate the following limit:

$\lim_{{h \to 0}} \frac{\sqrt{36 - x + h} - \sqrt{36 - x}}{h}$

This type of limit is a difference quotient, which suggests that we are looking for the derivative of a function. Specifically, this is the derivative of the function $f(x) = \sqrt{36 - x}$.

To solve this limit, we can apply the conjugate multiplication technique to eliminate the square roots in the numerator.

Step 1: Multiply by the conjugate

The conjugate of the numerator is $\sqrt{36 - x + h} + \sqrt{36 - x}$. We multiply both the numerator and denominator by this conjugate:

$\frac{\sqrt{36 - x + h} - \sqrt{36 - x}}{h} \cdot \frac{\sqrt{36 - x + h} + \sqrt{36 - x}}{\sqrt{36 - x + h} + \sqrt{36 - x}} = \frac{\left( \sqrt{36 - x + h} - \sqrt{36 - x} \right) \left( \sqrt{36 - x + h} + \sqrt{36 - x} \right)}{h \left( \sqrt{36 - x + h} + \sqrt{36 - x} \right)}$

Step 2: Simplify the numerator

Using the difference of squares formula:

$\left( \sqrt{36 - x + h} \right)^2 - \left( \sqrt{36 - x} \right)^2 = (36 - x + h) - (36 - x) = h$

So, the expression simplifies to:

$\frac{h}{h \left( \sqrt{36 - x + h} + \sqrt{36 - x} \right)} = \frac{1}{\sqrt{36 - x + h} + \sqrt{36 - x}}$

Step 3: Take the limit as $h \to 0$

As $h \to 0$, $\sqrt{36 - x + h} \to \sqrt{36 - x}$. Therefore, the denominator becomes:

$\sqrt{36 - x} + \sqrt{36 - x} = 2 \sqrt{36 - x}$

Thus, the limit becomes:

$\lim_{{h \to 0}} \frac{1}{2 \sqrt{36 - x}} = \frac{1}{2 \sqrt{36 - x}}$

Final Answer:

$\lim_{{h \to 0}} \frac{\sqrt{36 - x + h} - \sqrt{36 - x}}{h} = \frac{1}{2 \sqrt{36 - x}}$

This is the derivative of $f(x) = \sqrt{36 - x}$.

Would you like more details or have any questions?

Here are 5 related questions:

1. What is the general method for differentiating square root functions?
2. How is the conjugate method used in simplifying limits involving square roots?
3. Can you find the derivative of other square root functions, such as $f(x) = \sqrt{x^2 + 1}$?
4. How does this process relate to finding the slope of a curve at a point?
5. What are some other limit problems where multiplying by the conjugate is useful?

Tip: When dealing with square roots in limits, multiplying by the conjugate is a powerful tool for simplifying expressions.