#### Answer

$$\lim_{x\to\pi}\sin(x-\sin x)=0$$
The function is continuous at $x=\pi$

#### Work Step by Step

*Recall Theorem 10:
If $g$ is continuous at $b$ and $\lim_{x\to c}f(x)=b$, then $$\lim_{x\to c}g(f(x))=g(b)=g(\lim_{x\to c}f(x))$$
$$A=\lim_{x\to\pi}f(x)=\lim_{x\to\pi}\sin(x-\sin x)$$
For the domain $(-\infty,\infty)$, $\lim_{x\to c}\sin x= \sin c$
So the function $y=\sin x$ is continuous on $(-\infty,\infty)$
That means we can apply Theorem 10 here. In detail,
$$A=\sin(\lim_{x\to\pi} x-\lim_{x\to\pi}\sin x)$$
$$A=\sin(\pi-\sin\pi)$$
$$A=\sin(\pi-0)$$
$$A=\sin\pi=0$$
- Examine $f(\pi)$: $$\sin(\pi-\sin\pi)=\sin(\pi-0)=\sin\pi=0$$
As $\lim_{x\to\pi}f(x)=f(\pi)$, the function is continuous at $x=\pi$