How to show that $f: (0, \infty)\longrightarrow\Bbb R$ defined by $f(x)= 1/x$ is not Lipschitz continuous?

Question

How to show that $f: (0, \infty)\longrightarrow\Bbb R$ defined by $f(x)= 1/x$ is not Lipschitz continuous?

If $K$ is a Lipschitz constant, I got $$K\ge \frac{|f(x)-f(y)|}{|x-y|}=\frac1{|xy|}$$ and then I don't know what to do. I thought maybe $xy\to 0$ then there is no $K$ because $|1/xy|\to\infty$. Is it right?

Answer 1

By definition, $$ f\text{ Lipschitz}\implies |f(x)-f(y)|\le K|x-y|.$$ But take $x=1/n$, $y=1/2n$, $n\in\Bbb N$. $$|f(x)-f(y)|=|1/(1/n) - 1/(1/2n)|=n$$ And now...