Hi,

thank you very much for your kind help. Would you mind to say if my proof is ok?

Let $\displaystyle x=0$. Then $\displaystyle y<\epsilon^2 \Longrightarrow \sqrt{y}< \epsilon$

Let $\displaystyle x>y$

$\displaystyle |\sqrt{x}-\sqrt{y}|\leq\sqrt{|x-y|}$

$\displaystyle \sqrt{x}-\sqrt{y}\leq\sqrt{x-y}$

$\displaystyle \sqrt{x}\leq\sqrt{x-y}+\sqrt{y}$

$\displaystyle x\leq x-y+y+2\sqrt{\left(x-y\right)y}$

$\displaystyle x\leq x+2\sqrt{\left(x-y\right)y}$

In the case of y>x:

$\displaystyle |\sqrt{x}-\sqrt{y}|\leq\sqrt{|x-y|}$

$\displaystyle \sqrt{y}-\sqrt{x}\leq\sqrt{y-x}$

$\displaystyle \sqrt{y}\leq\sqrt{y-x}+\sqrt{x}$

$\displaystyle y\leq y-x+x+2\sqrt{\left(y-x\right)x}$

$\displaystyle y\leq y+2\sqrt{\left(y-x\right)x}$

$\displaystyle \delta=\epsilon^2$ because

$\displaystyle |x-y|<\epsilon^2 \Longrightarrow |\sqrt{x}-\sqrt{y}|\leq\sqrt{|x-y|}<\epsilon$

Is it ok?

Thanks