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$\sqrt{x}$ is defined over $\left [ 0,+\infty \right ]$

What's confusing me is that my teacher used to call the point $(0,0)$ of this function , a point of discontinuity, but this function is continuous over its domain

NoChance
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Pedro Alvarès
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    How do you define continuity of a function $f$ at a point $x$? – Klangen Jul 01 '19 at 08:31
  • Right and left hand limits are equal, and also equal to the image of the point of abscissa, let's say x. – Pedro Alvarès Jul 01 '19 at 08:38
  • What about the limit from the left at $x \to 0^-$? Also note that the derivative is not defined at $x=0$ – Yuriy S Jul 01 '19 at 08:41
  • And what is the limit from the left at $x=0$ equal? – Klangen Jul 01 '19 at 08:42
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    0 at the left is not inside the domain – Pedro Alvarès Jul 01 '19 at 08:45
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    @PedroAlvarès, there you have it. There might be different definitions by different authors, but since you claim that the function is not defined at $0^-$, then you can't claim that it's continuous at $x=0$, because you can't claim that the limits from the left and from the right are equal. – Yuriy S Jul 01 '19 at 08:57
  • @YuriyS: What should be adjusted is not the conclusion about continuity (because this functions is continuous without a doubt), but the incorrect definition “right and left hand limits are equal”. – Hans Lundmark Jul 01 '19 at 10:30
  • @HansLundmark, thank you for correcting me, I need to brush up on my basics – Yuriy S Jul 01 '19 at 10:52

4 Answers4

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I note the various answers here all discuss the continuity (which it is), but do not answer the question in the title - which is, what is the best name by which to call this point? To that end, I would say that the best name for it is probably the endpoint of the graph of $\sqrt{}$. And "graph" here is actually a strictly-defined mathematical object, not simply the thing you draw on paper, which is really a picture of the graph (cf. Rene Magritte's famous painting, The Treachery of Images). The graph of a function $f$ is the set of all ordered pairs $(x, y)$ such that $y = f(x)$, and can be considered as one of the three ingredients that define a function, the other two being the domain and codomain (though you could argue that only two are really required and one is redundant: the domain can be recovered from the graph, as the set of all the first elements of the ordered pairs).

In this case, the graph inherits a topology from the plane, and is topologically equivalent to the half-open interval $[0, \infty)$ (think of the curve as just a "bent half-line"). The point $0$ on that interval corresponds to $(0, 0)$ on the graph of $\sqrt{}$, hence since we call $0$ an endpoint of such interval, I'd call such an endpoint of the graph as well.

The_Sympathizer
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Let $f(x)=\sqrt{x}$ for $x \ge 0.$ Then:

$$ \lim_{x \to o}f(x)=0=f(0).$$

Consequence: $f$ is continuous at $0$.

Fred
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    But since $\lim_{x\to 0^-}\sqrt{x} = n$ for all $n\in\mathbb{R}$ (and in particular $0$ as well), wouldn't it rather be a point of discontinuity? – Klangen Jul 01 '19 at 08:44
  • Isn't there a name for (0,0) ? – Pedro Alvarès Jul 01 '19 at 08:46
  • 0 minus is negative – Pedro Alvarès Jul 01 '19 at 08:47
  • @PedroAlvarès: since the derivative of $\sqrt{x}$ blows up at $0$, I might call it a singular point. – robjohn Jul 01 '19 at 08:51
  • @Klangen What makes you state that $\lim_{x\to0^-}\sqrt{x}=n$ for all $n\in\mathbb R$? – drhab Jul 01 '19 at 08:52
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    @drhab By simple application of the definition of a limit. See Eric's answer here: https://math.stackexchange.com/questions/2811906/continuity-of-square-root-function-at-the-first-point – Klangen Jul 01 '19 at 08:53
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    @Klangen: saying that $\lim\limits_{x\to0^-}\sqrt{x}$ exists and $\lim\limits_{x\to0^-}\sqrt{x}=L$ for all $L\in\mathbb{R}$ is definitely non-standard. For instance, that would seem to imply that all $L\in\mathbb{R}$ are equal. – robjohn Jul 01 '19 at 09:00
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    @Klangen Mmmh... But looking at it like that we can prove things like $5=4$ because both equal $\lim_{x\to0^-}\sqrt x$. – drhab Jul 01 '19 at 09:01
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Topological answer.

If you are not familiar (yet) with topology then this answer might be quite useless for you. Nevertheless I publish it to promote completeness.

If $f$ is a function $X\to Y$ where domain and codomain are both underlying sets of topological spaces then $f$ is (according to the common definition in topology) continuous at $x_0\in X$ iff:

  • For every open set $U\subseteq Y$ with $f\left(x_0\right)\in U$ there is an open set $V\subseteq X$ such that $x_0\in V$ and $f\left(V\right)\subseteq U$

Applying that here on $X=[0,\infty)$ and $Y=\mathbb R$ and $f$ prescribed by $x\mapsto\sqrt{x}$ where the sets are equipped with their usual topologies we find that $f$ is continuous at $0$.

drhab
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$f$ is indeed continuous at $x=0$, as Fred already explained.

$f$ is not differentiable at 0, though, as $$\lim_{h \to 0^+} \frac{\sqrt{0+h}- \sqrt{0}}{h}$$ does not exist! Also, $f(x):=\sqrt{x}$, then $$\lim_{x \to 0} f'(x) = +\infty$$

If you are not studying pure mathematics, but rather physics or any other science, you teacher may have taken that license at call $x=0$ a "singularity", as $f$ is not "that smooth" there as it is in the rest of its domain

David
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