How can I prove $\frac{1}{x}$ is not Lipschitz on $(0,\infty)$?

Question

I know that first, this function is Lipschitz if we have some $c>0, |f(x)-f(y)| = |\frac{1}{x}-\frac{1}{y}| = \frac{|x-y|}{|xy|} \leq \frac{1}{c^2}|x-y|$ making it Lipschitz at least on $(c, \infty)$ but how can I show if it is $(0, \infty)$, this is no longer true? It doesn't make intuitive sense to me because as long as $\frac{1}{c^2}$ where c is positive (i.e. $(0, \infty) > 0)$, the function should still be Lipschitz since our original function does not change.

Answer 1

If $f$ was Lipschitz in $(0,\infty)$ then it would preserve Cauchy sequences. However, the Cauchy sequence $a_n=\frac{1}{n}$ is mapped by $f$ to the sequence $f(a_n)=n$, which is clearly not Cauchy.

Answer 2

Solution #1. A continuous and differentiable function is Lipshitz iff it has bounded derivative. But differentiating $\frac{1}{x}$ gives $-\frac{1}{x^2}$, which is clearly unbounded in $(0,\infty)$.

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Solution #2. Suppose it is Lipshcitz. As you have written we have $|f(x) - f(y)| = \frac{|x - y|}{|xy|} \leq M|x - y|$ for some $M > 0$. This implies that for all $x,y > 0$ such that $x \neq y$, we have $\frac{1}{|xy|} \leq M$, but this is clearly false as we can let, for instance, $x = \frac{1}{\sqrt{M}}$ and $y = \frac{1}{2\sqrt{M}}$.