Find all points of discontinuity of $f,$ where $f$ is defined by $f(x) = \begin{cases} \frac{|x|}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$

(D) The given function $f$ is $f(x) = \begin{cases} \frac{|x|}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$
It is known that $x < 0 \implies |x| = -x$ and $x > 0 \implies |x| = x$.
Therefore,the function can be rewritten as:
$f(x) = \begin{cases} -1, & \text{if } x < 0 \\ 0, & \text{if } x = 0 \\ 1, & \text{if } x > 0 \end{cases}$
The function $f$ is defined for all real numbers. Let $c$ be any real number.
Case $I$: If $c < 0$,then $f(c) = -1$.
$\lim_{x \to c} f(x) = \lim_{x \to c} (-1) = -1 = f(c)$.
Thus,$f$ is continuous for all $x < 0$.
Case $II$: If $c = 0$,the left-hand limit is $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-1) = -1$.
The right-hand limit is $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (1) = 1$.
Since the left-hand limit $\neq$ right-hand limit,$f$ is discontinuous at $x = 0$.
Case $III$: If $c > 0$,then $f(c) = 1$.
$\lim_{x \to c} f(x) = \lim_{x \to c} (1) = 1 = f(c)$.
Thus,$f$ is continuous for all $x > 0$.
Conclusion: $x = 0$ is the only point of discontinuity.