To evaluate the limit:

$\lim_{x \to 2} \frac{x^2 - 4}{x - 2},$

we notice that direct substitution of $x = 2$ leads to an indeterminate form $\frac{0}{0}$. To resolve this, we factorize the numerator $x^2 - 4$ as a difference of squares:

$x^2 - 4 = (x - 2)(x + 2).$

Substituting this factorization into the limit expression:

$\frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2}.$

For $x \neq 2$, the $(x - 2)$ terms cancel out, leaving:

\frac{x^2 - 4}{x - 2} = x + 2 \quad \text{(for x \neq 2)}.

Now, we evaluate the limit:

$\lim_{x \to 2} (x + 2) = 2 + 2 = 4.$

Final Answer:

$\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4.$

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Would you like further explanation or additional examples?

Here are 5 related questions for practice:

1. Evaluate $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$.
2. Find the limit: $\lim_{x \to 1} \frac{x^3 - 1}{x - 1}$.
3. Compute $\lim_{x \to a} \frac{x^2 - a^2}{x - a}$.
4. Determine the limit $\lim_{x \to 0} \frac{x^2 - 2x}{x}$.
5. Solve $\lim_{x \to -1} \frac{x^3 + 1}{x + 1}$.

Tip:

When encountering indeterminate forms, look for algebraic simplifications such as factoring, expanding, or using conjugates to simplify the expression.