# Show that the function $f(x)=\mid \sin x+\cos x\mid$ is continuous at $x=\pi$.

Toolbox:
• The composition of two continuous function is a continuous function.
• (i.e) $gof(x)=gof(a)$ at $x=a$
Step 1:
$f(x)=\mid \sin x+\cos x\mid$ at $x=\pi$
Let $g(x)=\sin x+\cos x$
$h(x)=\mid x\mid$
Hence $f(x)$ is a composite function .
(i.e) $f(x)=hog(x)$
Step 2:
Now let us prove that $f(x)$ is continuous at $x=\pi$
It is enough to prove that $g(x)$ is continuous at $x=\pi$ and $h(x)$ is continuous at $y=g(\pi)$
$g(\pi)=\sin\pi+\cos\pi$
$\qquad=-1$
$\lim\limits_{\large x\to \pi}g(x)=\lim\limits_{\large x\to \pi}(\sin x+\cos x)$
$\qquad\qquad=\sin\pi+\cos\pi$
$\qquad\qquad=-1$
Also $g(\pi)=-1$
Therefore $\lim\limits_{\large x\to \pi}g(x)=g(\pi)$
So,$g(x)$ is continuous at $x=\pi$
We have $g(\pi)=-1$
$\Rightarrow y=g(\pi)=-1$
Step 3:
Now ,
$\lim\limits_{\large y\to -1}h(y)=\lim\limits_{\large y\to -1}\mid y\mid$
$\qquad\qquad=\lim\limits_{\large y\to -1}-y$
$\qquad\qquad=-(-1)$
$\qquad\qquad=1$
$h(g(\pi))=h(-1)$
$\qquad\quad=\mid -1\mid$
$\qquad\quad=1$
Therefore $\lim\limits_{\large y\to 1}h(y)=h(g(\pi))$
$\Rightarrow \lim\limits_{\large g(x)\to 1}h(g(x))=h(g(\pi))$
$\Rightarrow \lim\limits_{\large g(x)\to g(\pi)}h(g(x))=h(g(\pi))$
(i.e) $h$ is continuous at $g(\pi)$
Hence $f(x)=hog(x)$ is continuous at $x=\pi$