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 $0/0$.