The problem asks to determine whether the function:

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

is continuous at $x = 2$, using the definition of continuity.

Step-by-step solution:

1. Simplify the function: The expression $x^2 - 4$ can be factored as $(x - 2)(x + 2)$, so the function becomes: $f(x) = \frac{(x - 2)(x + 2)}{x - 2}$ For $x \neq 2$, we can cancel $x - 2$ from both the numerator and the denominator, leaving: $f(x) = x + 2 \quad \text{for} \quad x \neq 2$

2. Definition of continuity: A function $f(x)$ is continuous at $x = 2$ if: $\lim_{x \to 2} f(x) = f(2)$ This means that the limit of the function as $x$ approaches 2 must equal the value of the function at $x = 2$.

3. Evaluate the limit: For $x \neq 2$, we have $f(x) = x + 2$. Now, take the limit as $x$ approaches 2: $\lim_{x \to 2} f(x) = \lim_{x \to 2} (x + 2) = 2 + 2 = 4$

4. Value of the function at $x = 2$: The original function $f(x) = \frac{x^2 - 4}{x - 2}$ is undefined at $x = 2$, since substituting $x = 2$ into the denominator gives zero. Therefore, $f(2)$ is not defined.

5. Conclusion: Since $f(2)$ is not defined, the function is not continuous at $x = 2$, even though the limit exists. The discontinuity is a removable discontinuity because the limit exists, but the function is not defined at that point.

Would you like more details or have any questions?

Here are some related questions to explore:

1. What is the precise definition of a removable discontinuity?
2. How can you modify this function to make it continuous at $x = 2$?
3. What is the difference between removable and non-removable discontinuities?
4. How can L'Hopital's Rule be applied to find limits of indeterminate forms like $\frac{0}{0}$?
5. What are other types of discontinuities besides removable?

Tip: When a function has a removable discontinuity, it can often be redefined to make it continuous by properly defining the function at that point.