Does the limit rule $\lim_{x \rightarrow 0}\frac{\sin x}{x}=1$ apply to $\lim_{x \rightarrow \pi}\frac{\sin\left(x-\pi\right)}{\left(x-\pi\right)}=1$? In my textbook, I was given an example below :
$$\lim_{x \rightarrow \pi}\frac{\sin\left(x-\pi\right)}{\left(x-\pi\right)}=1$$
Previously I was taught that this formula :
$$\lim_{x \rightarrow 0}\frac{\sin x}{x}=1$$
only applies when $x$ approaches $0$. Can someone explain to me? 

Yes of course and more in general for any $f(x)\to 0$ as $x\to x_0\in \mathbb{R}\cup\{\pm \infty\}$
$$\lim_{x \rightarrow x_0}\frac{\sin\left(f(x)\right)}{f(x)}=1$$
