$f(x)=\vert x \vert ^\frac{3}{2}$ is differentiable but not uniformly continuous 
Prove that $$f(x)=\vert x \vert ^{\large{\frac{3}{2}}}$$  is differentiable but not uniformly continuous  on $\Bbb R$

At $x \neq 0$,  $$f'(x)=\begin{cases} \frac{3}{2}\sqrt x&\text{if}\;x>0\\\\\frac{3}{2}\sqrt{-x}&\text{if}\;x<0 \end{cases}$$At $x=0$, $$f'(0)=\lim_{h \to 0} \frac{f(h)}{h}=0$$ so $$(\forall x \in \Bbb R): f'(x)=\frac{3}{2} \sqrt{\vert x \vert}$$
To prove  non uniform continuity using  sequential criterion, I pick $$(x_n=n )\wedge (y_n=n+\frac{1}{n})$$ but it does 't work.
Is the first one correct ? Any help for the second one?
 A: The function is not uniformly continuous on $\mathbb{R}$ because $\lim_{x \to \infty}f'(x) = +\infty$. 
Note that a bounded derivative is sufficient for uniform continuity but an unbounded derivative (when the limit does not exist) is not enough to conclude that continuity is not uniform on an unbounded domain. For example, $f(x) = \sin(x^3)/x$ is uniformly continuous on $(0,\infty)$ but has an unbounded derivative.
The proof uses the mean value theorem which gives $\xi \in (x,y)$ such that
$$|f(x) - f(y)| = |f'(\xi)||x-y|$$
For any $\delta > 0$, take $y = x + \delta/2$ so that $|x - y| = \delta/2 < \delta$ and
$$|f(x) - f(y)| = |f'(\xi)| \delta$$
If $x$ is sufficiently large, then the RHS is greater than $1$ since $f'(x) \to +\infty$.
A: The first part is fine. 
Contradict with $\epsilon = \frac 12$ the uniform continuity of $f$. Given any $\delta > 0$, we must show that there is a pair of points $x,y$ such that $|x-y |< \delta$ but $||x|^{\frac 32}  - |y|^{\frac 32}| > \frac 12$.
To do this, since $f'(x) \to \infty$ as $x \to \infty$, simply choose $R$ large enough so that $|x| > r \implies |f'(x)| > \frac{1}{\delta}$. Now, for any $x> r$ we have $|x - (x+0.5\delta)| < \delta$ but $|f(x) - f(x + 0.5\delta)| = 0.5 \delta \times |f'(y)|$ for some $y \in (x,x+0.5\delta)$ by the mean value theorem. However by choice we have $|f'(y)| > \frac{1}{\delta}$ so we get $|f(x) - f(x + 0.5\delta)| > \frac 12$. 
In general, any uniformly continuous function admits at most linear growth (one may check this via repeated applications of the triangle inequality). Since this function has superlinear growth, it is not uniformly continuous.
