Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x\to\pi}\sin(x+\sin x)=0$
$$A=\lim\limits_{x\to\pi}\sin(x+\sin x)$$ 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: $$A=f(\lim\limits_{x\to\pi}g(x))$$ $$A=\sin(\lim\limits_{x\to\pi}(x+\sin x))$$ $$A=\sin(\pi+\sin\pi)$$ $$A=\sin(\pi+0)$$ $$A=\sin\pi$$ $$A=0$$