Examine whether $f(x) = \sin |x|$ is a continuous function.

(N/A) Let $f(x) = \sin |x|$.
This function $f$ is defined for every real number and $f$ can be written as the composition of two functions $f = h \circ g$,where $g(x) = |x|$ and $h(x) = \sin x$.
$[\because (h \circ g)(x) = h(g(x)) = h(|x|) = \sin |x| = f(x)]$
First,we prove that $g(x) = |x|$ and $h(x) = \sin x$ are continuous functions.
$g(x) = |x|$ can be written as:
$g(x) = \begin{cases} -x, & \text{if } x < 0 \\ x, & \text{if } x \ge 0 \end{cases}$
Clearly,$g$ is defined for all real numbers. Let $c$ be a real number.
Case $I$: If $c < 0$,$g(c) = -c$ and $\lim_{x \to c} g(x) = \lim_{x \to c} (-x) = -c$. Thus,$\lim_{x \to c} g(x) = g(c)$. So,$g$ is continuous for $x < 0$.
Case $II$: If $c > 0$,$g(c) = c$ and $\lim_{x \to c} g(x) = \lim_{x \to c} (x) = c$. Thus,$\lim_{x \to c} g(x) = g(c)$. So,$g$ is continuous for $x > 0$.
Case $III$: If $c = 0$,$g(0) = 0$. $\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} (-x) = 0$ and $\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} (x) = 0$. Since $\lim_{x \to 0^-} g(x) = \lim_{x \to 0^+} g(x) = g(0)$,$g$ is continuous at $x = 0$.
Thus,$g(x) = |x|$ is continuous everywhere.
Now,$h(x) = \sin x$ is defined for every real number. Let $c$ be a real number. Put $x = c + k$. If $x \to c$,then $k \to 0$.
$\lim_{x \to c} h(x) = \lim_{k \to 0} \sin(c + k) = \lim_{k \to 0} (\sin c \cos k + \cos c \sin k) = \sin c(1) + \cos c(0) = \sin c = h(c)$.
Thus,$h(x) = \sin x$ is continuous everywhere.
Since the composition of two continuous functions is continuous,$f(x) = (h \circ g)(x) = \sin |x|$ is a continuous function.