How do I prove that $f : \Bbb{R}\setminus [2,4] → \Bbb{R}$ such that $x\mapsto \frac{1}{x-3}$ is Lipschitz continous?

Question

It sounds simple enough, but i how do i work with the interval? Do I have to apply a case distinction for $> [2,4]$ and $< [2,4]$?

Answer 1

$$|\frac{1}{x-3}-\frac{1}{y-3}|=|\frac{x-y}{(x-3)(y-3)}|\leq |x-y|\,,\forall x,y\in\Bbb{R}\setminus[2,4]$$