We are tasked with finding the limit $\lim_{a \to 9} f(a)$, where:

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

First, notice that directly substituting $a = 9$ into the expression gives a $\frac{0}{0}$ indeterminate form:

$f(9) = \frac{9 - 9}{3 - \sqrt{9}} = \frac{0}{0}$

Thus, we need to simplify the expression by removing the indeterminate form.

Step 1: Rationalize the denominator

The denominator has a square root, so we can rationalize it by multiplying both the numerator and denominator by the conjugate of the denominator, $3 + \sqrt{a}$:

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

Simplify the denominator using the difference of squares:

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

Now the expression becomes:

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

Step 2: Cancel out $9 - a$

Since $9 - a \neq 0$ for $a \neq 9$, we can cancel out $9 - a$ in the numerator and denominator:

$f(a) = 3 + \sqrt{a}, \quad \text{for} \ a \neq 9$

Thus, the continuous function that is equal to $f(a)$ everywhere except at $a = 9$ is:

$g(a) = 3 + \sqrt{a}$

Step 3: Evaluate the limit

Now that we have simplified the expression, we can evaluate the limit by directly substituting $a = 9$ into $g(a)$:

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

Final Answer:

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

Let me know if you would like more details or further clarification.

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Here are 5 related questions for further practice:

1. How would you handle $\lim_{a \to 0} \frac{a - 1}{\sqrt{a} - 1}$?
2. What is the limit of $\frac{\sqrt{x} - 2}{x - 4}$ as $x \to 4$?
3. How do you apply L'Hopital's rule to evaluate indeterminate forms?
4. What are conjugates, and how are they used in limit problems?
5. What is the importance of canceling common factors in rational functions?

Tip: Always check if simplifying the expression can remove an indeterminate form before attempting advanced methods like L'Hopital's rule.