Prove that $f: [0, \infty) \rightarrow \mathbb{R}, x \mapsto \sqrt{x}$ is consistently continuous

Question

Prove that $f: [0, \infty) \rightarrow \mathbb{R}, x \mapsto \sqrt{x}$ is consistently continuous

Let $\delta > 0$, let $\varepsilon > 0$, let $|x-x_{0}|< \delta$

$\Rightarrow$

$$|\sqrt{x}-\sqrt{x_{0}}| > |\sqrt{x-x_{0}}| < \sqrt{\delta} = \varepsilon$$

$$\delta = \varepsilon^{2}$$

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Is it correct?

Answer 1

Several things are wrong with this "proof," the most critical of which is the end statement $\delta = \epsilon^2$. If you start with just $\delta > 0$ and $\epsilon >0$ and come to the conclusion that $\delta = \epsilon^2$ you can be sure you did something wrong.

However, I see what you are doing: You are trying to find how small $\delta$ will have to be for any given $\epsilon$. Say you start with "Let $\delta = \frac14\epsilon^2$" and then proceed to say $$ | f(x) - f(x_0) | = |\sqrt{x}-\sqrt{x_0}| \leq \sqrt{|x-x_0|} = \sqrt{\delta} = \frac12 \epsilon < \epsilon $$ Well now you have something resembling a proof (though you should mention that this $\delta$ does not depend on $x_0$) but there is still a big step missing:

Why is $|\sqrt{x}-\sqrt{x_0}| \leq \sqrt{|x-x_0|}$ ????

That is the meaty part of the problem. The statement is true, but you need to prove it. (Hint: consider the case of $x>x_0$ and square both sides. Then consider other cases.)