Let's analyze the function $f(x) = \frac{x^2 - 4}{x - 2}$ and find where it is discontinuous.

1. Simplifying the function: The numerator $x^2 - 4$ is a difference of squares, so it can be factored: $f(x) = \frac{(x - 2)(x + 2)}{x - 2}$ For $x \neq 2$, we can cancel the $(x - 2)$ terms: $f(x) = x + 2 \quad \text{for} \quad x \neq 2$ However, when $x = 2$, the original expression is undefined because of division by zero. This means the function has a removable discontinuity at $x = 2$.

2. Evaluating the limit as $x \to 2$: Since $f(x) = x + 2$ for $x \neq 2$, we can find the limit as $x$ approaches 2: $\lim_{x \to 2} f(x) = \lim_{x \to 2} (x + 2) = 4$ Therefore, the limit exists at $x = 2$, but the function is not defined there.

3. Conclusion: The function is discontinuous at $x = 2$, with a removable discontinuity because the limit exists but the function is undefined at that point.

Based on this, the correct choice is likely A, where the function is discontinuous at the single value $x = 2$, and the limit is 4.

Would you like more details or have any questions about this? Here are some additional related questions you might find helpful:

1. What are the different types of discontinuities (removable, jump, infinite)?
2. How do you distinguish between a removable discontinuity and a non-removable one?
3. Can a function have more than one discontinuity point?
4. How does factoring help in identifying discontinuities?
5. What role do limits play in determining the behavior of functions at points of discontinuity?

Tip: Always check if a discontinuity is removable by simplifying the function, as factoring can reveal hidden behavior at certain points.