To solve this problem, let’s analyze the function $f(x) = \frac{x-2}{x^2-4}$ and find its discontinuities:

Step 1: Factor the denominator

The denominator $x^2 - 4$ is a difference of squares: $x^2 - 4 = (x - 2)(x + 2)$ Thus, the function becomes: $f(x) = \frac{x - 2}{(x - 2)(x + 2)}$

Step 2: Simplify the function

The term $x - 2$ in the numerator and denominator can cancel out, but this introduces a restriction: $f(x) = \frac{1}{x + 2}, \quad \text{for } x \neq 2$

Step 3: Identify discontinuities

The denominator is zero when $x + 2 = 0$, i.e., $x = -2$. Additionally, since $x - 2$ was canceled, $x = 2$ is also a point of discontinuity.

• At $x = 2$: This is a removable discontinuity, as canceling $x - 2$ removed it from the denominator.
• At $x = -2$: This is a nonremovable discontinuity, as the denominator remains zero even after simplification.

Final Answer:

• Removable discontinuities: $x = 2$
• Nonremovable discontinuities: $x = -2$

Would you like a more detailed explanation or clarification?
Here are 5 related questions for further exploration:

1. What is the graphical interpretation of removable and nonremovable discontinuities?
2. How would the behavior of $f(x)$ differ near $x = -2$ versus $x = 2$?
3. Can all rational functions have removable discontinuities?
4. How do limits relate to the classification of discontinuities?
5. What would happen if $f(x)$ was modified to include $x = 2$ explicitly?

Tip: When simplifying rational functions, always check for restrictions introduced by canceled terms.