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If the function is continuous at some point, then is it necessary that limit should exists at that point?

Like in case of $\sqrt{x}$, its continuous at $x=0$, but limit doesn't exist as left hand limit is not defined.

But it appears meaningless to say that function is continuous at some point, but limit doesn't exist at that point.

Any inputs?

user3290550
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  • If you have a function defined on a closed interval, then you can only talk about one-sided limits at the endpoints of that interval: the function is not defined on the "other" side of each of those endpoints. – NickD Dec 13 '19 at 02:16
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    Related: https://math.stackexchange.com/questions/637280/limit-of-sqrt-x-as-x-approaches-0. Also note that you could have more extreme examples, like $f(x)=\sqrt{x^2 (x^2-1)}$ which is continuous according to the $\epsilon$-$\delta$ definition (including at $x=0$, where the limit doesn't exist since that's an isolated point in the domain $D_f$). – Hans Lundmark Dec 13 '19 at 03:46
  • See also https://math.stackexchange.com/questions/455296/can-a-function-with-just-one-point-in-its-domain-be-continuous – Hans Lundmark Dec 13 '19 at 03:48
  • Under the more general definition of limit (see the linked questions), the limit will exist at $c$ if $f$ is continuous at $c$ unless $c$ is an isolated point of the domain. – bjorn93 Dec 13 '19 at 13:38
  • what about if a function is just one point, in that case does limit exists at that point?

    Can we say function is continuous at that point?

     – user3290550 Dec 14 '19 at 11:09

2 Answers2

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Note that $f(x)=\sqrt{x}$ is defined on $[0,\infty)$, to stay accord with the continuity-limit property, we talk about \begin{align*} \lim_{x\rightarrow 0,x\in[0,\infty)}f(x), \end{align*} which is essentially the right-sided limit of $f$ at $0$. The running variable must at least stay in the domain of the function, or else it just doesn't make sense.

user284331
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If you're familiar with $\varepsilon-\delta$ arguments:

Recall that we say a function $f$ is continuous at a point $x_0$ if for all $\varepsilon>0$, there exists $\delta>0$ such that if $|x-x_0|<\delta$, then $|f(x)-f(x_0)|<\varepsilon$. For the function $f$ given by $f(x)=\sqrt x$, no such $\delta$ can exist, since for any negative $x$ a distance less than $\delta$ from $x_0$, $f(x)$ is undefined. So $f$ is not continuous at $0$.

We can however say that $f$ is right-continuous at $0$, or equivalently, for all $\varepsilon>0$ there exists $\delta>0$ such that if $0\leq x-x_0<\delta$, then $|f(x)-f(x_0)|<\varepsilon$. Then $\sqrt x$ satisfies the definition of right-continuity at the origin. Recalling the definition of the right-sided limit, you can convince yourself that $f$ is right-continuous at $x_0$ if and only if the limit as $x\to x_0^+$ exists.

csch2
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  • what about if a function is just one point, in that case does limit exists at that point?

    Can we say function is continuous at that point?

     – user3290550 Dec 14 '19 at 11:09
  • @user3290550 often we say that for a limit to be defined at a point, it must be a cluster point of the domain, so for any $\varepsilon>0$ there must be points of the domain within a $\varepsilon$-neighborhood of the point you’re taking the limit at. So according to the definition I’m used to then we can’t say anything about limits or continuity at isolated points. – csch2 Dec 14 '19 at 17:03