[Math] Show that $\sqrt{x}$ is uniformly continuous but not Lipschitz.

Question

I'm having problems proving the function $f(x)=\sqrt{x}$ is uniformly continuous, but not a Lipschitz function on the interval $(0,\infty)$. If I assume U.C , I can show it's not Lipschitz. However, I'm having problems showing it's U.C on that interval.

Answer 1

On the closed interval $[0,1]$, and hence on $(0,1]$, we have uniform continuity because of continuity. On $[1,\infty)$ we have uniform continuity because the derivative is bounded.