How to decide the neighbourhood when proving a function is locally Lipschitz-continuous. I have come across a problem in one of my textbooks that I understand but could not replicate.
Show that the function $f : \mathbb{R}^+ → \mathbb{R}$ where $f(x) = {1\over x}$ is locally Lipschitz-continuous in $\mathbb{R}^+ = (0, \infty)$.
$$|f(y) − f(z)| = |{1\over y}−{1\over z}|={|y − z|\over yz}≤{|y − z|\over (x/2)^2}$$
where it chooses the neighbourhood of U  $\subset$ $\mathbb{R}^+$ as U =$ ({x\over 2}, 2x)$ proving that it is locally Lipschitz-continuous in $\mathbb{R}^+ = (0, \infty)$ with $L = {4\over x^2}$
I am very confused on how they choose this value of U, if anyone could explain that would be amazing. 

Fact: If the derivative of a function is bounded over an interval, then the function is Lip on that interval with bound equal to the supremum of the derivatie there. 
Derivative of $1/x$ is $1/x^2$ which is bounded from above as long as we're away from zero. So, in fact $1/x$ is Lipschitz over any $[\delta, +\infty)$, $\delta >0$.
