Showing continuity with $\epsilon,\delta$ definition

Question

Let $f:(0,+\infty)\rightarrow\mathbb{R}$, $x\mapsto \sqrt{x}$ be a function.

I want to show that $f$ is continuous on $(0,+\infty)$

My approach:

Let $\epsilon>0$ and $x,x_{0}\in(0,+\infty)$

$|f(x)-f(x_{0})|=|\sqrt{x}-\sqrt{x_{0}}|=\left|\frac{x-x_{0}}{\sqrt{x}+\sqrt{x_{0}}}\right|=\frac{|x-x_{0}|}{\sqrt{x}+\sqrt{x_{0}}}<\frac{|x-x_{0}|}{\sqrt{x_{0}}}$

If $|x-x_{0}|<\delta$

Then we can write: $|f(x)-f(x_{0})|<\frac{\delta}{\sqrt{x_{0}}}$

Let $\delta=\epsilon\sqrt{x_{0}}$

Then we have $|f(x)-f(x_{0})|<\frac{\epsilon\sqrt{x_{0}}}{\sqrt{x_{0}}}=\epsilon$

Edit: corrected a mistake

Answer 1

Perhaps, one should take $\delta=\min\{|x_{0}|/2,(2^{-1/2}+1)\sqrt{x_{0}}\epsilon\}$, then $|x|=|x-x_{0}+x_{0}|\geq|x_{0}|-|x-x_{0}|>|x_{0}|-\delta>|x_{0}|-|x_{0}|/2=|x_{0}|/2$, so $\left|\dfrac{x-x_{0}}{\sqrt{x}+\sqrt{x_{0}}}\right|<\dfrac{\delta}{(2^{-1/2}+1)\sqrt{x_{0}}}<\epsilon$.

Answer 2

No, your proof is not correct.

Note that the following inequality $$ |x|=|x-x_{0}+x_{0}|\ge||x-x_{0}|+x_{0}|$$is not valid.

For example if $x=1$ and $x_0=10$ you will get $|x|=1$ while $ ||x-x_{0}|+x_{0}|= 19 $