#### Answer

$\lim\limits_{x\to\pi}\sin(x+\sin x)=0$

#### Work Step by Step

$$A=\lim\limits_{x\to\pi}\sin(x+\sin x)$$
Let $f(x)=\sin x$ and $g(x)=x+\sin x$, we have $$A=\lim\limits_{x\to\pi}f(g(x))$$
Since $\sin x$ is continuous on $R$, we can apply Theorem 8: $$A=f(\lim\limits_{x\to\pi}g(x))$$
$$A=\sin(\lim\limits_{x\to\pi}(x+\sin x))$$
$$A=\sin(\pi+\sin\pi)$$
$$A=\sin(\pi+0)$$
$$A=\sin\pi$$
$$A=0$$