Discuss the continuity of the function $f$ given by
$f(x) = \begin{cases} x, & \text{if } x \ge 0 \\ x^2, & \text{if } x < 0 \end{cases}$

(N/A) Clearly,the function is defined at every real number. The graph of the function is shown in the figure. By inspection,it is prudent to partition the domain of definition of $f$ into three disjoint subsets of the real line.
Let $D_1 = \{ x \in \mathbb{R} : x < 0 \}$,$D_2 = \{ 0 \}$,and $D_3 = \{ x \in \mathbb{R} : x > 0 \}$.
Case $1$: At any point in $D_1$,we have $f(x) = x^2$,which is a polynomial function and is continuous everywhere in its domain.
Case $2$: At any point in $D_3$,we have $f(x) = x$,which is a polynomial function and is continuous everywhere in its domain.
Case $3$: Now we analyze the function at $x = 0$. The value of the function at $0$ is $f(0) = 0$.
The left-hand limit of $f$ at $0$ is:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x^2 = 0^2 = 0$.
The right-hand limit of $f$ at $0$ is:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0$.
Thus,$\lim_{x \to 0} f(x) = 0 = f(0)$,and hence $f$ is continuous at $x = 0$. Since $f$ is continuous at every point in its domain,$f$ is a continuous function.