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? 

Set $y=x-\pi$. Then $y$ approaches $0$ if and only if $x$ approaches $\pi$. So we may write the following 
$$\lim_{x \rightarrow \pi}\frac{\sin\left(x-\pi\right)}{\left(x-\pi\right)}=\lim_{y \rightarrow 0}\frac{\sin y}{y}=1$$
EDIT: One may also see it as a composition of limits.
