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The limit expression you provided is:

\[
\lim_{{x \to 9}} \frac{{\sqrt{x} - \sqrt{9}}}{{x - 9}}
\]

To solve this, let's follow these steps:

### Step 1: Simplify the expression.
We recognize that \(\sqrt{9} = 3\), so the expression becomes:

\[
\lim_{{x \to 9}} \frac{{\sqrt{x} - 3}}{{x - 9}}
\]

### Step 2: Use the conjugate to simplify.
To simplify this expression, we multiply both the numerator and denominator by the conjugate of the numerator, which is \(\sqrt{x} + 3\):

\[
\frac{{\sqrt{x} - 3}}{{x - 9}} \times \frac{{\sqrt{x} + 3}}{{\sqrt{x} + 3}} = \frac{{(\sqrt{x} - 3)(\sqrt{x} + 3)}}{{(x - 9)(\sqrt{x} + 3)}}
\]

### Step 3: Simplify the numerator.
Using the difference of squares:

\[
(\sqrt{x} - 3)(\sqrt{x} + 3) = x - 9
\]

Now the expression simplifies to:

\[
\frac{{x - 9}}{{(x - 9)(\sqrt{x} + 3)}} = \frac{1}{{\sqrt{x} + 3}}
\]

### Step 4: Take the limit.
Now we can substitute \(x = 9\) directly into the expression:

\[
\lim_{{x \to 9}} \frac{1}{{\sqrt{9} + 3}} = \frac{1}{{3 + 3}} = \frac{1}{6}
\]

### Final Answer:
\[
\lim_{{x \to 9}} \frac{{\sqrt{x} - \sqrt{9}}}{{x - 9}} = \frac{1}{6}
\]

Would you like more details or have any other questions?

Here are five related questions for further practice:
1. What is the limit of \(\lim_{{x \to 16}} \frac{{\sqrt{x} - 4}}{{x - 16}}\)?
2. How do you handle limits that result in indeterminate forms like \(\frac{0}{0}\)?
3. Can limits always be evaluated using direct substitution, or when should you apply other techniques?
4. What other algebraic techniques can be used to simplify limits besides multiplying by the conjugate?
5. How can L'Hopital's Rule be used to solve limits of indeterminate forms?

**Tip:** When dealing with square roots in limits, multiplying by the conjugate is a powerful technique to simplify the expression and resolve indeterminate forms like \(\frac{0}{0}\).

Evaluate the limit \(\lim_{{x \to 9}} \frac{{\sqrt{x} - \sqrt{9}}}{{x - 9}}\).

This problem demonstrates how to evaluate the limit \(\lim_{{x \to 9}} \frac{{\sqrt{x} - 3}}{{x - 9}}\) using algebraic techniques such as rationalization and applying limit laws. Suitable for students in Grades 10-12.

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