The question asks to evaluate the limit:

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

Step-by-Step Solution:

1. Simplify the numerator: The numerator $x^2 - 4$ is a difference of squares. Thus: $x^2 - 4 = (x - 2)(x + 2).$

2. Rewrite the fraction: Substituting the simplified form into the fraction: $\frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2}.$

3. Cancel out the common factor: Since $x \neq 2$ (as we are approaching 2, not evaluating directly), we can cancel $x - 2$: $\frac{(x - 2)(x + 2)}{x - 2} = x + 2.$

4. Evaluate the limit: Now, take the limit as $x \to 2$: $\lim_{x \to 2} (x + 2) = 2 + 2 = 4.$

Final Answer:

$\boxed{4}$

Would you like additional explanations or have further questions?

Related Questions:

1. What are the key conditions for canceling terms in limits?
2. How does factoring simplify the computation of limits with indeterminate forms?
3. Can L'Hôpital's Rule be used to solve this problem? Why or why not?
4. What are some common patterns in factoring for polynomial expressions?
5. How do you evaluate a one-sided limit compared to a two-sided limit?

Tip:

Always check if the numerator and denominator can be simplified when solving limits that result in indeterminate forms like $0/0$.