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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.

Find the limit as x approaches 3 of the expression (sqrt(x^2 - 5) - 2) / (2x - 6).

Learn how to find the limit of the expression (sqrt(x^2 - 5) - 2) / (2x - 6) as x approaches 3, using algebraic techniques like rationalization and factoring. This problem covers key calculus concepts and is suitable for high school students in Grades 11-12.

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