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

(N/A) The function $f$ is defined at all points of the real line.
Case $1$: If $c < 1$,then $f(c) = c + 2$. Therefore,$\lim_{x \to c} f(x) = \lim_{x \to c} (x + 2) = c + 2 = f(c)$.
Thus,$f$ is continuous at all real numbers less than $1$.
Case $2$: If $c > 1$,then $f(c) = c - 2$. Therefore,$\lim_{x \to c} f(x) = \lim_{x \to c} (x - 2) = c - 2 = f(c)$.
Thus,$f$ is continuous at all points $x > 1$.
Case $3$: If $c = 1$,then the left-hand limit of $f$ at $x = 1$ is
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x + 2) = 1 + 2 = 3$.
The right-hand limit of $f$ at $x = 1$ is
$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x - 2) = 1 - 2 = -1$.
Since the left-hand limit and right-hand limit 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 $f$.