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 $[a, b]$.

Consider the interval $[0,1]$, then clearly $f(x) = x^{1/2}$ is continuous.
Can you show that $f$ is not Lipschitz on $[0,1]$? (Hint: Use the Mean-Value theorem on a closed sub-interval of $(0,1/n]$)
