We are asked to find $\lim_{a \to 9} f(a)$, where $f(a) = \frac{9 - a}{3 - \sqrt{a}}$.

First, note that substituting $a = 9$ directly into the function results in an indeterminate form $\frac{0}{0}$. Therefore, we need to simplify $f(a)$ by finding a continuous function that matches $f(a)$ everywhere except at $a = 9$.

Step 1: Simplify the expression

We begin by simplifying the expression $f(a) = \frac{9 - a}{3 - \sqrt{a}}$. To do this, we multiply both the numerator and the denominator by the conjugate of the denominator:

$f(a) = \frac{9 - a}{3 - \sqrt{a}} \times \frac{3 + \sqrt{a}}{3 + \sqrt{a}} = \frac{(9 - a)(3 + \sqrt{a})}{(3 - \sqrt{a})(3 + \sqrt{a})}.$

Step 2: Simplify the denominator

The denominator simplifies using the difference of squares formula:

$(3 - \sqrt{a})(3 + \sqrt{a}) = 9 - a.$

Thus, we now have:

$f(a) = \frac{(9 - a)(3 + \sqrt{a})}{9 - a}.$

Step 3: Cancel the common factor

As long as $a \neq 9$, we can cancel the factor $9 - a$ in both the numerator and the denominator, giving:

$f(a) = 3 + \sqrt{a}.$

Step 4: Compute the limit

Now, $f(a) = 3 + \sqrt{a}$ is continuous at $a = 9$. So, we can evaluate the limit by direct substitution:

$\lim_{a \to 9} f(a) = 3 + \sqrt{9} = 3 + 3 = 6.$

Thus, the limit is:

$\lim_{a \to 9} f(a) = 6.$

Would you like more details, or do you have any questions?

Here are five related questions to further explore this topic:

1. How do we know when to use the conjugate method to simplify limits?
2. What other indeterminate forms require simplification techniques?
3. Can L'Hopital's Rule be applied to solve this limit instead?
4. How do we check if a function is continuous at a point?
5. What is the general process for solving limits involving radicals?

Tip: When simplifying functions with square roots, multiplying by the conjugate is a useful technique to eliminate the radical and simplify the expression.