Question

Select the correct answer.

Answer 1

Answer:

Answer A.

Step-by-step explanation:

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

we will calculate the lateral limits of f when x approches x=2. Note that

$\lim_{x\to 2^{+}} \frac{x^2-4}{x-2} = \lim_{x\to 2^{+}} \frac{(x-2)(x+2)}{x-2} = 2+2 = 4$

$\lim_{x\to 2^{-}} \frac{x^2-4}{x-2} = \lim_{x\to 2^{-}} \frac{(x-2)(x+2)}{x-2} = 2+2 = 4$

We can clasify the discontinuity as follows:

- Removable discontinuity if both lateral limits are equal and finite.

- Jump discontinuity if both lateral limits are finite but different.

- Essential discontinuity if one of the limits is not finite and the other one is finite.

Based on this classification, since both lateral limits are equal, the discontinuity is a removable discontinuity