The given limit is:

\[
\lim_{x \to 2} \frac{x^2 - 4}{x - 2}.
\]

### Step 1: Simplify the expression
The numerator \(x^2 - 4\) is a difference of squares, so we can factorize it as:
\[
x^2 - 4 = (x - 2)(x + 2).
\]

Substitute this back into the expression:
\[
\frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2}.
\]

### Step 2: Cancel common terms
For \(x \neq 2\), the \(x - 2\) terms cancel out, leaving:
\[
\frac{(x - 2)(x + 2)}{x - 2} = x + 2.
\]

### Step 3: Evaluate the limit
Since the simplified expression is \(x + 2\), the limit as \(x \to 2\) is:
\[
\lim_{x \to 2} (x + 2) = 2 + 2 = 4.
\]

### Final Answer:
\[
\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4.
\]

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Would you like further details or have any questions? Here are some related questions for further exploration:

1. What happens if you don't cancel the common factor and directly substitute \(x = 2\)?
2. How does this relate to the concept of removable discontinuities?
3. Can you describe why this limit exists despite the original function being undefined at \(x = 2\)?
4. What would the limit be if \(x^2 - 4\) were replaced with another polynomial like \(x^2 - 5x + 6\)?
5. How does factoring simplify the process of finding limits in rational functions?

**Tip:** Always check if the numerator and denominator can be factored to simplify the evaluation of limits, especially when direct substitution results in an indeterminate form like \(0/0\).

Explore the step-by-step process of evaluating the limit lim_{x \to 2} (x^2 - 4)/(x-2). Learn key concepts such as factoring polynomials, simplifying rational expressions, and applying limit laws to solve this problem effectively. Suitable for high school students in Grades 10-12.

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