Show that the function defined by $f(x)=|\cos x|$ is a continuous function.

(N/A) The given function is $f(x)=|\cos x|$.
This function $f$ is defined for every real number and can be written as the composition of two functions $f=g \circ h$,where $g(x)=|x|$ and $h(x)=\cos x$.
$[\because (g \circ h)(x) = g(h(x)) = g(\cos x) = |\cos x| = f(x)]$
First,we prove that $g(x)=|x|$ and $h(x)=\cos x$ 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 $c < 0$,then $g(c)=-c$ and $\lim_{x \to c} g(x) = \lim_{x \to c} (-x) = -c = g(c)$. Thus,$g$ is continuous for $x < 0$.
Case $II$: If $c > 0$,then $g(c)=c$ and $\lim_{x \to c} g(x) = \lim_{x \to c} x = c = g(c)$. Thus,$g$ is continuous for $x > 0$.
Case $III$: If $c=0$,then $g(0)=0$. $\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} (-x) = 0$ 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,$h(x)=\cos x$ is defined for every real number. Let $c$ be a real number. Put $x=c+h$. As $x \to c$,$h \to 0$.
$\lim_{x \to c} h(x) = \lim_{h \to 0} \cos(c+h) = \lim_{h \to 0} (\cos c \cos h - \sin c \sin h) = \cos c(1) - \sin c(0) = \cos c = h(c)$.
Thus,$h(x)=\cos x$ is continuous everywhere.
Since the composition of two continuous functions is continuous,$f(x) = (g \circ h)(x) = |\cos x|$ is a continuous function.