We are tasked with estimating the limit:

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

Step-by-Step Solution:

1. Direct Substitution: First, let's check what happens if we directly substitute $x = 9$: $\frac{\sqrt{9} - 3}{9 - 9} = \frac{3 - 3}{0} = \frac{0}{0}$ This gives an indeterminate form $\frac{0}{0}$. Therefore, we need to simplify the expression further.

2. Rationalizing the Numerator: We can handle this indeterminate form by multiplying the numerator and denominator by the conjugate of the numerator: $\frac{\sqrt{x} - 3}{x - 9} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3}$ This will simplify the numerator using the difference of squares: $\frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{(x - 9)(\sqrt{x} + 3)} = \frac{x - 9}{(x - 9)(\sqrt{x} + 3)}$ Now, cancel the $x - 9$ terms (assuming $x \neq 9$): $\frac{1}{\sqrt{x} + 3}$

3. Substituting $x = 9$: Now we can directly substitute $x = 9$ into the simplified expression: $\frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}$

Thus, the limit is:

$\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9} = \frac{1}{6}$

Would you like further details or clarification?

Here are 5 related questions to expand your understanding:

1. How does rationalizing the numerator help in simplifying limits with indeterminate forms?
2. What other methods can be used to resolve indeterminate forms besides rationalization?
3. How would this limit change if we altered the square root term in the original function?
4. Can you find another example where the indeterminate form $\frac{0}{0}$ is resolved by simplifying the function?
5. How does this approach relate to the concept of continuity in calculus?

Tip: Always check for indeterminate forms first when evaluating limits, and try algebraic manipulation such as factoring or rationalizing to simplify the expression.