Does the limit rule $\lim_{x \rightarrow 0}\frac{\sin x}{x}=1$ apply to $\lim_{x \rightarrow \pi}\frac{\sin\left(x-\pi\right)}{\left(x-\pi\right)}=1$?

Question

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?

Answer 1

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.

Answer 2

Yes of course and more in general for any $f(x)\to 0$ as $x\to x_0\in \mathbb{R}\cup\{\pm \infty\}$

$$\lim_{x \rightarrow x_0}\frac{\sin\left(f(x)\right)}{f(x)}=1$$

Answer 3

You can use L'Hospital's rule to verify: $$\lim_{x \rightarrow \pi}\frac{\sin\left(x-\pi\right)}{\left(x-\pi\right)}=\lim_{x \rightarrow \pi}\frac{\cos\left(x-\pi\right)}{1}=\cos0=1.$$