$\sqrt{x}$ isn't Lipschitz function A function f such that 
$$
|f(x)-f(y)| \leq C|x-y|
$$
for all $x$ and $y$, where $C$ is a constant independent of $x$ and $y$, is called a Lipschitz function
show that $f(x)=\sqrt{x}\hspace{3mm} \forall x \in \mathbb{R_{+}}$ isn't Lipschitz function
Indeed,  there is no such constant C where 
$$
|\sqrt{x}-\sqrt{y}| \leq C|x-y| \hspace{4mm} \forall x,y \in \mathbb{R_{+}}
$$
we have only that inequality
$$
|\sqrt{x}-\sqrt{y}|\leq |\sqrt{x}|+|\sqrt{y}|
$$
Am i right ?
remark for @Vintarel  i plot it i don't know  graphically "Lipschitz" mean? 
what is the big deal in the graph of the square-root function 

in wikipedia they said 
Continuous functions that are not (globally) Lipschitz continuous The function f(x) = $\sqrt{x}$ defined on [0, 1] is not Lipschitz continuous. This function becomes infinitely steep as x approaches 0 since its derivative becomes infinite. However, it is uniformly continuous as well as Hölder continuous of class $C^{0,\alpha}$, α for $α ≤ 1/2$.
Reference
1] could someone explain to me this by math and not by words, please ??
2] what does "Lipschitz" mean graphically?

You have
$${\sqrt{1/n} - \sqrt{0}\over{1/n - 0}} = {1/\sqrt{n}\over {1\over n}} = \sqrt{n}.$$
This ratio can be made as large as you like by choosing $n$ large.  Therefore the square-root function fails to be Lipschitz.
