Show there is a closed interval $[a, b]$ such that the function $f(x) = |x|^{\frac1{2}}$ is continuous but not Lipschitz on on $[a, b]$. Hi guys I was given this as an "exercise" in my calculus class and we weren't told what a Lipschitz is so i really need some help, heres the question again:  
Show there is a closed interval $[a, b]$ such that the function $f(x) = |x|^{\frac1{2}}$ is continuous but not Lipschitz on on $[a, b]$.

Suppose $f$ is Lipscitz on $[0, 1]$. Then there is a positive constant $L$ such that $|f(x)-f(y)\leq L|x-y|$ for all $x$, $y\in[0, 1]$. In particular, for $y=0$ we have $\sqrt{x}=|f(x)-f(0)\leq L|x|$. This implies that $\displaystyle\lim_{x\rightarrow 0^+}\frac{\sqrt{x}}{x}=\displaystyle\lim_{x\rightarrow 0^+}\frac{1}{\sqrt{x}}=\infty$. We arrive at contradiction.
