We need to evaluate the limit:

$\lim_{x \to 3} \frac{x - 3}{\sqrt{x^2 - 9}}$

Step 1: Substitute $x = 3$ directly into the expression

Substituting $x = 3$ directly:

$\frac{3 - 3}{\sqrt{3^2 - 9}} = \frac{0}{\sqrt{9 - 9}} = \frac{0}{0}$

We get the indeterminate form $\frac{0}{0}$, so we need to simplify the expression further.

Step 2: Factor and simplify the denominator

Notice that $x^2 - 9$ is a difference of squares, so we can factor it as:

$x^2 - 9 = (x - 3)(x + 3)$

Now, the limit becomes:

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

Step 3: Simplify the expression

We can cancel the common factor of $x - 3$ from the numerator and denominator. However, before canceling, we should express the denominator as:

$\sqrt{(x - 3)(x + 3)} = \sqrt{x - 3} \cdot \sqrt{x + 3}$

Thus, the limit now becomes:

$\frac{1}{\sqrt{x + 3}}$

Step 4: Evaluate the limit

Now we can substitute $x = 3$ directly into the simplified expression:

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

Thus, the value of the limit is:

$\frac{1}{\sqrt{6}}$

Final Answer:

$\lim_{x \to 3} \frac{x - 3}{\sqrt{x^2 - 9}} = \frac{1}{\sqrt{6}}$

Would you like any additional details or further steps explained? Here are a few related questions:

1. How can we use L'Hopital's Rule to solve limits with indeterminate forms like $\frac{0}{0}$?
2. What is the general method for handling square roots in limits?
3. Can this limit be generalized for other values of $x$ approaching a number other than 3?
4. How do we handle limits involving square roots when the expression contains both a numerator and denominator with square roots?
5. What does the difference of squares formula tell us about limits and factorizations?

Tip: Always check for indeterminate forms like $\frac{0}{0}$ when approaching a limit. If encountered, try factoring, simplifying, or applying L'Hopital's Rule.