We are tasked with finding the limit of the following expression as $x \to 3$:

$\lim_{x \to 3} \frac{\sqrt{x^2 - 5} - 2}{2x - 6}.$

Step 1: Substituting $x = 3$

First, let's substitute $x = 3$ directly into the expression to check if we can evaluate it directly:

$\frac{\sqrt{3^2 - 5} - 2}{2(3) - 6} = \frac{\sqrt{9 - 5} - 2}{6 - 6} = \frac{\sqrt{4} - 2}{0} = \frac{2 - 2}{0} = \frac{0}{0}.$

This results in an indeterminate form $\frac{0}{0}$, so we need to simplify further using algebraic techniques.

Step 2: Simplifying by Rationalizing the Numerator

We will rationalize the numerator by multiplying the expression by the conjugate of the numerator, which is $\sqrt{x^2 - 5} + 2$. This will help eliminate the square root.

$\frac{\sqrt{x^2 - 5} - 2}{2x - 6} \times \frac{\sqrt{x^2 - 5} + 2}{\sqrt{x^2 - 5} + 2} = \frac{(\sqrt{x^2 - 5})^2 - 2^2}{(2x - 6)(\sqrt{x^2 - 5} + 2)}.$

Simplifying the numerator:

$(\sqrt{x^2 - 5})^2 - 2^2 = (x^2 - 5) - 4 = x^2 - 9.$

So the expression becomes:

$\frac{x^2 - 9}{(2x - 6)(\sqrt{x^2 - 5} + 2)}.$

Step 3: Factor and Cancel

Notice that $x^2 - 9$ is a difference of squares:

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

Also, $2x - 6$ can be factored as $2(x - 3)$. Now the expression becomes:

$\frac{(x - 3)(x + 3)}{2(x - 3)(\sqrt{x^2 - 5} + 2)}.$

We can cancel out the $(x - 3)$ terms:

$\frac{x + 3}{2(\sqrt{x^2 - 5} + 2)}.$

Step 4: Substitute $x = 3$ Again

Now, substitute $x = 3$ into the simplified expression:

$\frac{3 + 3}{2(\sqrt{3^2 - 5} + 2)} = \frac{6}{2(\sqrt{9 - 5} + 2)} = \frac{6}{2(2 + 2)} = \frac{6}{2 \times 4} = \frac{6}{8} = \frac{3}{4}.$

Final Answer:

The limit is $\frac{3}{4}$, which corresponds to option B.

Do you want further clarifications on any step?

Here are five related questions to explore this further:

1. How do you determine when to use conjugates to simplify a limit problem?
2. What are other common ways to handle the indeterminate form $\frac{0}{0}$?
3. Can you explain the difference of squares formula in more depth?
4. How would you handle limits involving higher powers of $x$ in similar problems?
5. What other indeterminate forms exist, and how can you resolve them?

Tip: Always check for simple algebraic manipulations, like factoring or rationalizing, when dealing with indeterminate forms in limits.