The given function is:

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

We need to find the values of $x = a$ where the function is discontinuous, and for each value of $a$, determine the limit of the function as $x$ approaches $a$. We also need to note if the limit does not exist.

Step 1: Simplify the function

The function $f(x) = \frac{x^2 - 4}{x - 2}$ contains a quadratic expression in the numerator. We can factor the numerator:

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

Thus, the function becomes:

$f(x) = \frac{(x - 2)(x + 2)}{x - 2}$

For $x \neq 2$, we can cancel the $(x - 2)$ terms, and the simplified function is:

$f(x) = x + 2 \quad \text{for} \quad x \neq 2$

Step 2: Identify discontinuity

The function is simplified to $f(x) = x + 2$, but there is still a discontinuity at $x = 2$ because the original function had the factor $\frac{0}{0}$ at $x = 2$. Therefore, $f(x)$ is discontinuous at $x = 2$.

Step 3: Determine the limit as $x$ approaches $2$

We now compute the limit of $f(x)$ as $x$ approaches $2$. Since the simplified function is $f(x) = x + 2$ for $x \neq 2$, we can compute the limit as:

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

Thus, the limit of $f(x)$ as $x$ approaches $2$ is $4$.

Conclusion:

• The function is discontinuous at $x = 2$.
• The limit of the function as $x$ approaches $2$ is $4$, but the function is not defined at $x = 2$ itself, causing the discontinuity.

Would you like more details or explanations about any specific steps?

Related Questions:

1. What is the formal definition of a removable discontinuity?
2. Can we make the function continuous at $x = 2$? How?
3. How would the behavior of the function change if the denominator was $x - 3$?
4. What happens to the limit of the function as $x \to \infty$?
5. How would this problem change if there was an additional term like $\frac{1}{x-2}$ added to the function?

Tip: Always check for factors that can be canceled to simplify rational functions before evaluating limits.