We are tasked with evaluating the limit:

$\lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3}$

Step-by-Step Solution:

1. Direct Substitution:

   • If we substitute $x = 9$ directly into the expression, we get:

   $\frac{9 - 9}{\sqrt{9} - 3} = \frac{0}{3 - 3} = \frac{0}{0}$

   This results in an indeterminate form, $\frac{0}{0}$, so we need to manipulate the expression further.

2. Factorization/Algebraic Manipulation:

   • Notice that we can manipulate the denominator $\sqrt{x} - 3$. Since we have a square root, let's multiply both the numerator and denominator by the conjugate of the denominator $\sqrt{x} + 3$ to simplify it.

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

3. Simplifying the Expression:

   • The denominator becomes a difference of squares:

   $(\sqrt{x} - 3)(\sqrt{x} + 3) = (\sqrt{x})^2 - 3^2 = x - 9$

   So, the expression simplifies to:

   $\frac{(x - 9)(\sqrt{x} + 3)}{x - 9}$

   • We can cancel out $x - 9$ from both the numerator and the denominator (as long as $x \neq 9$):

   $\lim_{x \to 9} \sqrt{x} + 3$

4. Final Calculation:

   • Now, substitute $x = 9$ into the simplified expression:

   $\sqrt{9} + 3 = 3 + 3 = 6$

Thus, the limit is $\boxed{6}$.

Would you like more details or have any questions?

Here are 5 related questions to explore this topic further:

1. How do we handle limits that result in the indeterminate form $\frac{0}{0}$?
2. Why is multiplying by the conjugate a useful technique in simplifying limits involving square roots?
3. What other techniques can we use to solve limits besides factoring or multiplying by the conjugate?
4. What are the conditions under which we can cancel terms in limits?
5. How does L'Hôpital's Rule help in solving indeterminate forms like $\frac{0}{0}$?

Tip: Always check if the expression simplifies easily before resorting to advanced techniques like L'Hôpital's Rule.