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please check my proof

we have

$|x-0|<\delta \rightarrow |\sqrt{x}-\sqrt{0}|<\epsilon $

$\rightarrow \sqrt{x}<\epsilon $

$\rightarrow x<\epsilon ^{2}$

choose $\delta=\epsilon ^{2}$

therefore $|x-0|<\delta \rightarrow |\sqrt{x}-\sqrt{0}|<\sqrt{\epsilon ^{2}}=\epsilon $

Then it is continuous at $ 0 $

Lingnoi401
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2 Answers2

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Maybe interesting to prove something stronger ...

For all $(x,y)\in\mathbb{R}^2$ such $0\le x\le y$ :

$$\sqrt y\le\sqrt{y-x}+\sqrt x$$

(obvious by comparing the squares)

So, for all $(x,y)\in[0,+\infty)^2$ :

$$|\sqrt y-\sqrt x|\le\sqrt{|y-x|}$$

Given $\epsilon>0$, the condition $|y-x|\le\epsilon^2$ implies $|\sqrt y-\sqrt x|\le\epsilon$.

This proves that $t\mapsto\sqrt t$ is uniformly continuous, hence continuous, hence continuous at 0.

Adren
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  • It's just asked proof for continuous at a point but that thank for example for uniformly continuous How about my proof Is it correct? – Lingnoi401 Mar 17 '17 at 07:51
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    @Lingnoi401: Your reasoning is correct, but IMHO badly written. You should emphasize that it is a proof by sufficient condition (and replace "we have" at the top by "we would like to show that for any $\epsilon>0$ there exists $\delta>0$ such that ..."). And also you should use "\implies" to get $\implies$ (instead of $\rightarrow$). – Adren Mar 17 '17 at 08:07
  • thank you for comment ^ ^ – Lingnoi401 Mar 17 '17 at 08:12
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Your use of $\leftrightarrow$ is awful !

Let $ \epsilon >0$. Put $\delta =\epsilon^2$. If $0 \le x < \delta$ then $0 \le\sqrt{x}< \epsilon$

Fred
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