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The given limit is:

\[
\lim_{x \to 6} \frac{\sqrt{x + 3} - 3}{x - 6}
\]

To evaluate this limit, let's use algebraic manipulation by multiplying the numerator and denominator by the conjugate of the numerator to eliminate the square root. 

1. Multiply the expression by \(\frac{\sqrt{x + 3} + 3}{\sqrt{x + 3} + 3}\):

\[
\frac{\sqrt{x + 3} - 3}{x - 6} \cdot \frac{\sqrt{x + 3} + 3}{\sqrt{x + 3} + 3} = \frac{(\sqrt{x + 3} - 3)(\sqrt{x + 3} + 3)}{(x - 6)(\sqrt{x + 3} + 3)}
\]

2. Simplify the numerator using the difference of squares:

\[
(\sqrt{x + 3} - 3)(\sqrt{x + 3} + 3) = (x + 3) - 9 = x - 6
\]

Now, the expression simplifies to:

\[
\frac{x - 6}{(x - 6)(\sqrt{x + 3} + 3)}
\]

3. Cancel the common factor \((x - 6)\):

\[
\frac{1}{\sqrt{x + 3} + 3}
\]

4. Substitute \(x = 6\):

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

Therefore, the limit is:

\[
\boxed{\frac{1}{6}}
\]

Would you like further details or have any questions? Here are some related problems to explore:

1. What happens if the limit is taken as \(x \to 0\) instead of \(x \to 6\)?
2. How can you evaluate limits involving other types of indeterminate forms?
3. How would the result change if the expression were \(\sqrt{x - 3}\) instead of \(\sqrt{x + 3}\)?
4. What are other methods to simplify limits involving square roots?
5. How can L'Hopital's Rule be applied to this problem?

**Tip:** Multiplying by the conjugate is a common technique to simplify expressions with square roots when direct substitution leads to an indeterminate form.

Compute \( \lim_{{x \to 6}} \frac{{\sqrt{x + 3} - 3}}{{x - 6}} \).

Explore a detailed step-by-step guide on how to compute the limit as x approaches 6 for the function \( \frac{{\sqrt{x + 3} - 3}}{{x - 6}} \). This guide covers essential algebraic techniques including the use of the difference of squares and simplification of indeterminate forms, suitable for high school students in Grades 11-12.

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