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Is $f(x) = \sqrt{x}$ is continuous at $0$?

I see that people say that $\sqrt{x}$ is continuous at the interval $[0,\infty)$. Their proof is based on that: (continuous from the right): $$\lim_{x \to 0^+}f(x) = f(0) = 0$$ At least from what I have seen: Is function continuous at 0?

But there is no left limit, namely $\lim_{x \to 0^-}$ doesnt exists.

And from what I know the statement is that: $f$ is continuous at $x_0$ if and only if, it continuous from the left and from the right.

But here, there is no limit from the left.

So how $\sqrt{x}$ is continuous at $0$ so we say it continuous at $[0,\infty)?$

dmtri
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Alon
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    Common misconception. If the function is not defined on the left, the limit is the right-hand limit. You always work in the domain. –  Dec 27 '19 at 12:12
  • Ok, i didnt know that, so why it is an if and only if statment? wierd – Alon Dec 27 '19 at 12:21
  • See: https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit#Precise_statement_for_real_valued_functions

    The epsilon-delta is the "strict" definition. To test for continuity, you have to see if the epsilon-delta definition works for sqrt(x). Note that in the wikipedia article, x must be in the domain D.

     – Adam Rubinson Dec 27 '19 at 12:23
  • Please show the reference where you found "$f$ is continuous at $x_0$ if and only if, it continuous from the left and from the right." –  Dec 27 '19 at 12:57
  • This is written in my book in front of me, but it is in hebrew, i looked at the internet and realy found what you said. maybe i miss understand something in the book. but its exactly the translation. – Alon Dec 27 '19 at 13:03

3 Answers3

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Since the function is not defined on the left of 0 then the notion of continuity in 0 consists only of the continuity at the right of 0.

theX
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Everyone will agree that the function $g: \Bbb R \to \Bbb R$

$$ g(x) = \left\{\begin{array}{lr} 0, & \text{for } x \lt 0 \\ \sqrt x , & \text{for } x \ge 0 \end{array}\right\} $$

is continuous.

Restrict this function to the domain $[0,+\infty)$, defining

$\tag 1 \displaystyle{f = g_{\;|\,[0,+\infty)}}$

We leave it to OP to come up with sensible terminology/definitions concerning continuity for the function $f$. But if you want the restriction of a continuous function to also be continuous (seems reasonable), you won't have much 'leg room'.

CopyPasteIt
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There is no thing as "continuity from the right (or left)". Referring to limits, a function $f\colon G\to\mathbb R$ is continuous in $x\in G$ iff for every sequence $(x_n)$ with elements in $G$ that converges to $x$ we can conclude that $\lim_{n\to\infty}f(x_n)=f(x)$. Therefore, e.g., if $G=\mathbb Z$, we have that $f$ is continuous on $G$.

Michael Hoppe
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