We are asked to find the limit of the following function as $x$ approaches 3 from the right:

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

Step 1: Analyze the function

First, let's express the function in a simpler form:

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

Notice that the denominator involves $\sqrt{x^2 - 9}$, which is equivalent to:

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

So, we can rewrite the function as:

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

Step 2: Examine the behavior as $x \to 3^+$

As $x$ approaches 3 from the right, we get the following:

1. $x - 3$ approaches 0.
2. $x + 3$ approaches 6.
3. The square root $\sqrt{(x - 3)(x + 3)}$ approaches $\sqrt{0 \times 6} = 0$.

Thus, both the numerator and denominator approach 0, so this is an indeterminate form of type $\frac{0}{0}$, and we can apply algebraic manipulation to simplify the expression.

Step 3: Simplify the expression

Since we have a factor of $(x - 3)$ in both the numerator and denominator, we can cancel one occurrence of $(x - 3)$. To do this, notice:

$f(x) = \frac{x - 3}{\sqrt{(x - 3)(x + 3)}} = \frac{1}{\sqrt{x + 3}}$

Step 4: Evaluate the limit

Now, evaluate the limit of the simplified expression as $x \to 3^+$:

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

Final Answer:

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

Let me know if you'd like more details or have any questions! Here's a list of related questions:

1. How would the limit behave if $x$ approaches 3 from the left?
2. What happens to the function if $x = 3$?
3. How can we handle indeterminate forms like $\frac{0}{0}$ using L'Hôpital's Rule?
4. How does simplifying the square root expression help in calculating limits?
5. How can this technique be applied to other similar limits?

Tip: When dealing with square roots in limits, try to factor or simplify the expression before applying the limit, especially when it leads to indeterminate forms!