$\sqrt{x}$ 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?
 A: Suppose that $\sqrt{x}$ is a Lipschitz function, then there exists $C$ such that 
$$\Big|\frac{\sqrt{y}-\sqrt{x}}{y-x}\Big| \le C$$
Now, Let $y=2x$, so
$$(\sqrt{2}-1)x^{-\frac{1}{2}}\le C$$
Letting $x→0$ gives a contradiction.
A: $x \geq y$ implies $\sqrt{x} \geq \sqrt{y}$ (monotonous)
Then you need $\sqrt{x} - \sqrt{y} \leq C(x-y)$ for $x \ge y$
Use $a^2 - b^2 = \left(a-b\right)\left(a+b\right)$ to divide both sides by the (positive) $\sqrt{x} - \sqrt{y}$ to get $1 \leq C(\sqrt{x} + \sqrt{y})$, or $C \geq \frac{1}{\sqrt{x} + \sqrt{y}}$. Obviously the frac diverges as $(x,y)$ approaches $(0,0)$ so there is no upper bound $C$ to satisfy the requirement.
A: Hint: why is it not possible to find a $C$ such that
$$
|\sqrt{x} - \sqrt{0}|\leq C|x-0|
$$
For all $x \geq 0$?
As a general rule: Note that a differentiable function will necessarily be Lipschitz on any interval on which its derivative is bounded.
In response to the wikipedia excerpt: "This function becomes infinitely steep as $x$ approaches $0$" is another way of saying that $f'(x) \to \infty$ as $x \to 0$. If you look at slope of the tangent line at each $x$ as $x$ gets closer to $0$,  those tangent lines become steeper and steeper, approaching a vertical tangent at $x = 0$.
"Graphically", we can say that a differentiable function will be Lipschitz (if and) only if it never has a vertical tangent line.
Some functions that are not Lipschitz due to an unbounded derivative:
$$
f(x) = x^{1/3}\\
f(x) = x^{1/n},\quad n = 2,3,4,5,\dots
$$
A more subtle example:
$$
f(x) = x^2,\quad x \in \mathbb{R}\\
f(x) = \sin(x^2), \quad x \in \mathbb{R}
$$
Note in these cases that although $f'(x)$ is continuous, there is no upper bound for $f'(x)$ over the domain of interest.
A: $\sqrt{}$ is monotonous, so just assume $x \geq y$, then you can drop the absolute values and it simplifies to $1 \leq C(\sqrt{x} + \sqrt{y}$. Since you can make the sum of square roots arbitrarily small (by suitably decreasing $x$ and $y$), as soon as it's smaller than $1/C$ the inequality no longer holds.
A: Showing $\sqrt {x}$ is not Lipschitz .
We prove this by contradiction method.
Suppose if possible $\sqrt x$ is Lipschitz function. Then there exists $C>0$ such that
$|√x-√y|<C|x-y|$ for all $x,y\in \mathbb R^{+}$.
$\\$
Take $x=(\frac{1}{1+C})^2$ and $y=0$.
Then $|\frac{1}{1+C}|<C((\frac{1}{1+C})^2)$
$\Rightarrow C+1\leq C$ which is not possible.
So $√x$ is not Lipschitz function.
A: 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.
