Question

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

Answer 1

Let's assume that $h\left(x\right)=\cos \left(x\right)$ and $g\left(x\right)=|x|$

So $f\left(x\right)=\left(goh\right)=|\cos \left(x\right)|$

=> Let's prove that $h\left(x\right)=\cos \left(x\right)$ is continuous in it's domain.

Let $c$ be a real number, put $x=c+h$

So if $x=c$ then it means that $h\to 0$

$h\left(c\right)=\cos \left(c\right)$

$\underset{x\to c}{lim}$ $h\left(x\right)$ = $\underset{x\to c}{lim}$ $\cos \left(x\right)$

Put $x=h+c$

And as mentioned above, when $x\to c$ then it means that $h\to 0$

Which gives us $\underset{h\to 0}{lim}$ $cos(c+h)$

Expanding $\cos (c+h)$ = $\cos \left(c\right)\cos \left(h\right)-\sin \left(c\right)\sin \left(h\right)$

Which gives us $\underset{h\to 0}{lim}$ $\cos \left(h\right)\cos \left(c\right)-\sin \left(h\right)\sin \left(c\right)$

$=\cos \left(c\right)\cos \left(0\right)-\sin \left(c\right)\sin \left(0\right)$

$=\cos \left(c\right)$

This gives us

$\underset{x\to c}{lim}$ $h\left(x\right)$ = $\underset{x\to c}{lim}$ $\cos \left(x\right)$ = $\cos \left(c\right)=h\left(c\right)$

And this proves that $\cos \left(x\right)$ is continuous all across its domain

=> Here $g\left(x\right)$ is given by $|x|$

And $|x|=-x$ for $x<0$ and $x$ for $x>0$

For $c<0$

$\underset{x\to c}{lim}g\left(x\right)=-c=g\left(c\right)$

and

For $c>0$

$\underset{x\to c}{lim}g\left(x\right)=c=g\left(c\right)$

which shows that $|x|$ is continuous in its domain

=> So by theorem: If function $h$ and function $g$ are continuous then $goh$ is also continuous.

$\therefore$ $|\cos \left(x\right)|$ is continuous across its domain.