# 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

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 $$x_0>\frac{1}{2}$$

• Can you just use $x_0=\frac{1}{2}$ which will reduce the calculation? You overkilled it. Dec 6, 2022 at 9:06
• Also it doesn’t work for $x_0=\frac{1}{2}$ since $\delta=0$ Dec 6, 2022 at 9:09
• true that. sorry my fault, but for $x>\frac{1}{2}$ it should be correct?!?
– jaki
Dec 6, 2022 at 9:14
• To my eyes it seems correct. For $x_0=\frac{1}{2}$ do the same thing then the $\sqrt{2x_0-1}$ term will vanish and you'll get it. Dec 6, 2022 at 9:20
• Um... so which is it? Do you have to prove $f$ is continuos and $x =\frac 12$? or for all $x$ so that $x > \frac 12$? Dec 6, 2022 at 9:37

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

• It's not correct, $(2x-1)/\sqrt{2x-1} < 2x - 1$ is true only if $\sqrt{2x-1} > 1$. Dec 6, 2022 at 9:52
• As I said, it's incorrect, jaki. Also, this is not an answer to your question since you changed the question for no good reason. I'm not sure why are you so keen on overcomplicating stuff while you simply have to solve inequality $\sqrt{2x-1} < \varepsilon$. Dec 6, 2022 at 10:05
• jaki, by definition it is precisely enough to show that all $x\in (1/2,1/2+\delta)$ satisfy the inequality. So pick $\delta$ so they do. Dec 6, 2022 at 10:19
• Changed it again. Hope it is correct now.
– jaki
Dec 6, 2022 at 10:41
• It is correct now. Dec 6, 2022 at 14:15