$\text { Discuss the continuity of the function } f ( x ) = \frac { x ^ { 2 } - 4 } { x - 2 } \text {. }$
A) $f ( x )$ is discontinuous at $x = - 2$.
B) $f ( x )$ is discontinuous at $x = - 2,2$.
C) $f ( x )$ is discontinuous at $x = 2$.
D) $f ( x )$ is continuous for all real $x$.
E) $f ( x )$ is continuous at $x = 4$.

Question

$\text { Discuss the continuity of the function } f ( x ) = \frac { x ^ { 2 } - 4 } { x - 2 } \text {. }$ 
A) $f ( x )$ is discontinuous at $x = - 2$. 
B) $f ( x )$ is discontinuous at $x = - 2,2$. 
C) $f ( x )$ is discontinuous at $x = 2$. 
D) $f ( x )$ is continuous for all real $x$. 
E) $f ( x )$ is continuous at $x = 4$.

Quizplus · Accepted Answer

The function has a removable singularity at x = 2. We can simplify the function as follows:
$$f(x) = \frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2}= x+2, \ 	ext{ for } x
eq 2.$$ 
Thus, the limit of the function as x approaches 2 is 4. However, at x = 2, the function is undefined since we can't divide by zero. Therefore, the function is discontinuous at x = 2. It is continuous everywhere else.

Discuss the continuity of the function f(x)=x2−4x−2. \text { Discuss the continuity of the function } f ( x ) = \frac { x ^ { 2 } - 4 } { x - 2 } \text {. } Discuss the continuity of the function f(x)=x−2x2−4​.

$\text { Discuss the continuity of the function } f ( x ) = \frac { x ^ { 2 } - 4 } { x - 2 } \text {. }$