Prove continuity of $f: [\frac{1}{2},\infty ) \rightarrow \mathbb{R}: x \mapsto \sqrt{2x-1}$ for $x_0>\frac{1}{2}$ with Epsilon-delta definition

Question

Prove continuity of $f: [\frac{1}{2},\infty ) \rightarrow \mathbb{R}: x \mapsto \sqrt{2x-1}$ for $x_0>\frac{1}{2}$ with Epsilon-delta definition of continuity

show:
$\forall \epsilon >0 \ \exists \delta>0 \ \forall x \in [\frac{1}{2}, \infty): ( \ |x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|<\epsilon \ )$

$|x-x_0|<\delta$

$|\sqrt{2x-1} - \sqrt{2x_0-1} | = |\frac{(2x-1)-(2x_0-1)}{\sqrt{2x-1} + \sqrt{2x_0-1}}| = |\frac{2(x-x_0)}{\sqrt{2x-1} + \sqrt{2x_0-1}}|=\frac{2|x-x_0|}{\sqrt{2x-1} + \sqrt{2x_0-1}}$

$\sqrt{2x-1}>0 \Rightarrow \frac{2|x-x_0|}{\sqrt{2x-1} + \sqrt{2x_0-1}}<\frac{2|x-x_0|}{\sqrt{2x_0-1}} \Rightarrow |f(x)-f(x_0)|<\frac{2\delta}{\sqrt{2x_0-1}}$

let $\delta=\frac{\epsilon \sqrt{2x_0-1}}{2} \Rightarrow |f(x)-f(x_0)|<\frac{\epsilon \sqrt{2x_0-1}}{\sqrt{2x_0-1}}=\epsilon$

Which shows $\forall \epsilon >0 \ \exists \delta>0 \ \forall x \in [\frac{1}{2}, \infty): ( \ |x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|<\epsilon \ )$
and therefore the continuity of $f(x)$ for $x_0>\frac{1}{2}$ is proven.

Is this prove correct?

Edit: changed $x=\frac{1}{2}$ to $x_0>\frac{1}{2}$

Answer 1

For $x_0=\frac{1}{2}$:

$|f(x)-f(\frac{1}{2})|=|f(x)|=\left|\sqrt{2x-1}\right|<\epsilon$

$\forall x \in ( \frac{1}{2},\frac{1}{2}+\delta): \left|\sqrt{2x-1}\right| \stackrel{!}{<} \epsilon$

choose: $\delta:= \frac{\epsilon^{2}}{2} \Rightarrow \forall x \in ( \frac{1}{2},\frac{1}{2}+\delta): \left|\sqrt{2x-1}\right| < \epsilon$

And therefore the continuity of $f(x)$ in $x_0=\frac{1}{2}$ is proven