## University Calculus: Early Transcendentals (3rd Edition)

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