Find the points of discontinuity of $f,$ where $f(x) = \begin{cases} \frac{\sin x}{x}, & \text{if } x < 0 \\ x + 1, & \text{if } x \ge 0 \end{cases}$

(NONE) The given function $f$ is $f(x) = \begin{cases} \frac{\sin x}{x}, & \text{if } x < 0 \\ x + 1, & \text{if } x \ge 0 \end{cases}$
It is evident that $f$ is defined at all points of the real line.
Let $c$ be a real number.
Case $I$: If $c < 0,$ then $f(c) = \frac{\sin c}{c}$ and $\lim_{x \to c} f(x) = \lim_{x \to c} \left( \frac{\sin x}{x} \right) = \frac{\sin c}{c}.$
Since $\lim_{x \to c} f(x) = f(c),$ $f$ is continuous at all points $x < 0.$
Case $II$: If $c > 0,$ then $f(c) = c + 1$ and $\lim_{x \to c} f(x) = \lim_{x \to c} (x + 1) = c + 1.$
Since $\lim_{x \to c} f(x) = f(c),$ $f$ is continuous at all points $x > 0.$
Case $III$: If $c = 0,$ then $f(0) = 0 + 1 = 1.$
The left-hand limit at $x = 0$ is $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin x}{x} = 1.$
The right-hand limit at $x = 0$ is $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x + 1) = 1.$
Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 1,$ $f$ is continuous at $x = 0.$
From the above observations,$f$ is continuous at all points of the real line. Thus,$f$ has no points of discontinuity.