Lipschitz and uniform continuity Show that f(x)=$\sqrt{x}$ is uniformly continuous, but not Lipschitz continuous.
I can prove that it's uniformly continuous. But why is it not Lipschits? How do I check the definition?

Given any $M>0$, choose $c=0$ and $ 0<x<\frac{1}{m^{2}}$, so that $\frac{1}{\sqrt{x}}>M$. then we have:
$$ \frac{|f(x)-f(c)|}{|x-c|}=\frac{|\sqrt{x}|}{|x|}=\frac{1}{\sqrt{x}}>M $$
Since $M$ was arbitrary, this shows that $F$ is not Lipschitz continuous. 
