Discuss the continuity of the function $f$ defined by $f(x) = \begin{cases} x + 2, & \text{if } x \leq 1 \\x - 2, & \text{if } x > 1 \end{cases}$.

Discontinuous at $x=1$

The correct answer is Option (2) → Discontinuous at $x=1$ ##

The function $f$ is defined at all points of the real line.

$\textbf{Case 1}$ If $c < 1$, then $f(c) = c + 2$. Therefore, $\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} (x + 2) = c + 2$

Thus, $f$ is continuous at all real numbers less than 1.

$\textbf{Case 2}$ If $c > 1$, then $f(c) = c - 2$. Therefore,

$\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} (x - 2) = c - 2 = f(c)$

Thus, $f$ is continuous at all points $x > 1$.

$\textbf{Case 3}$ If $c = 1$, then the left hand limit of $f$ at $x = 1$ is

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (x + 2) = 1 + 2 = 3$

The right hand limit of $f$ at $x = 1$ is

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} (x - 2) = 1 - 2 = -1$

Since the left and right hand limits of $f$ at $x = 1$ do not coincide, $f$ is not continuous at $x = 1$. Hence $x = 1$ is the only point of discontinuity of $f$.