Prove that $\exp (-\frac{1}{x})$ is Lipschitz continuous on $(0,\infty)$.
My idea for any $x,y > 0$, we know that $\max\{\exp (-\frac{1}{x}), \exp (-\frac{1}{y})\}\leq 1 $
Could I simply choose my Lipschitz constant as $L:=2$, since
$\lvert \exp (-\frac{1}{x})- \exp (-\frac{1}{y})\rvert \leq 2$ by the triangle inequality? I then however have no dependency on $x,y$. Any ideas?
As, $f$ is continuous and differentiable on $(0,\infty) $.
So, using mean value theorem on $f$ over $[x,y]$, $|f(x)-f(y)| = |f'(z)| |x-y| $ for some $z\in (x,y) $, where $x,y$ arbitrarily choosen from $(0,\infty) $
Now as, $f'(x)= \frac{e^{-\frac{1}{x}}}{x^2} $ ,it is bounded on $(0,\infty) $,
So, , $|f(x)-f(y)| \le |f'(z)| |x-y| \le M |x-y| $