Is $h(x)=x^\frac{3}{2}$, $D_h=[0,\infty)$ uniformly continuous?

Question

Is $h(x)=x^\frac{3}{2}$, $D_h=[0,\infty)$ uniformly continuous ?

My attempt:

i) since $h$ is continuous on $[0,2]$ then it is uniformly continuous on $[0,2]$.

ii) If $x,u \in [1,\infty)$, then

$|h(x)-h(u)|=|x^\frac{3}{2} - u^\frac{3}{2} |$

$=\frac{|(x-u)||(x^2+xu+u^2)|}{| x^\frac{3}{2} +u^\frac{3}{2} |}$

$\leq \frac{|(x-u)|(|x|^2+|xu|+|u|^2)|}{| x^\frac{3}{2} +u^\frac{3}{2} |} $

I want to prove that $\frac{|(x-u)|(|x|^2+|xu|+|u|^2)|}{| x^\frac{3}{2} +u^\frac{3}{2} |} \leq a |x-u|$

Then take $\delta = \frac{\epsilon }{a}$ to get that

$\forall x,u \in [1,\infty), |x-u|<\delta \implies |h(x)-h(u)|<\epsilon$

How can I complete please?

Answer 1

It is not uniformly continuous. Take the sequences $a_n=(n+1)^{\frac{2}{3}}$ and $b_n=n^{\frac{2}{3}}$. Then $\lim_{n\to\infty} (a_n-b_n)=0$ but $\lim_{n\to\infty} (h(a_n)-h(b_n))\ne 0$.