# Doubts: continuity VS existence of the limit

I'm having some doubts that I cannot solve about this "conundrum", if I can call it like so.

Consider $$f(x) = \sqrt{x}$$.

Now we all know that $$\sqrt{0} = 0$$, hence the square root is well defined at $$x = 0$$.

$$f$$ is continuous at $$x = 0$$ because $$\lim_{x\to 0} \sqrt{x} \equiv f(0) = 0$$. But now I was thinking: the limit at $$x = x_0$$ exists if and only if

$$\lim_{x\to x_0^+} f(x) = \lim_{x\to x_0^-} f(x)$$

Yet here $$\lim_{x\to 0^-} \sqrt{x}$$ doesn't exist.

How do we solve this "conundrum"? How can we talk about continuity at $$x = 0$$ in this case?

• A simpler approach to this issue is to define continuity in the topological context (or metric space context), so the role of the domain is clear. This kind of definition is biased toward functions defined on an open interval around every point. Nov 30, 2023 at 14:27
• A possible issue here is that Calculus textbooks often give definitions that only apply in a limited setting. So your book might define continuity at $x_0$ only when $x_0$ is in the interior of the domain, and then their definition doesn't apply here. If they define it more generally, then the theorem that you're using (about one-sided limits) doesn't apply (unless you put in an extra hypothesis, such as that $x_0$ is in the interior of the domain, just for this theorem). Sometimes the book gets mixed up and states a theorem that doesn't work with the definition that they used! Nov 30, 2023 at 14:36
• Read the definitions in your textbook carefully. Not all textbooks agree on this (they have "simplified" definitions intended for beginners). The answers here seem to assume they know the definition; maybe they agree with your textbook, maybe they don't. Nov 30, 2023 at 16:01
• Does this answer your question? Is a function defined at a single point continuous?
– Jam
Dec 4, 2023 at 13:07

## 4 Answers

Not an expert in real analysis here, but when checking for continuity you only approach the function from where it exists.

Let $$f$$ be a function whose domain and range are sets of real numbers. Let $$f$$ be defined at $$x_0$$. If these conditions are met, then $$f$$ is continuous at $$x_0$$ if: $$(\forall\varepsilon>0)(\exists\delta>0)(\forall x\overbrace{\in\text{dom}f}^\text{key part})(|x-x_0|<\delta\implies|f(x)-f(x_0)|<\varepsilon)$$.

The key part is where it says $$x$$ must belong to the domain of the function as you approach it. Only when a function is defined on some region to both the left AND right of the point you're looking at does the theorem on two sided limits apply. Two sided limits are a special case.

Good question. Limits are weird.

You might take this to the extreme and ask "What if the whole function is one isolated point in the $$x$$-$$y$$ plane? Then you can't approach it from the left or the right." The surprising answer is that a single point is continuous according to the rigorous definition of continuity (after all, it doesn't exactly have any jumps).

• That was enlightening, thank you! Nov 30, 2023 at 14:32

The domain of the square root is $$[0,∞)$$, so you cannot approach $$0$$ from below. It therefore suffices to approach $$0$$ from above! The limit should be the same no matter how one approaches $$0$$, and in this case there is only one direction to do it from.

• So in this case the continuity is like "right continuity"? I mean in the definition I would take only $0^+$? Nov 30, 2023 at 14:19
• Yes, if the point in question (here $0$) lies on the end of the domain of the function, this is what one usually means with continuity at the point. You should of course keep this in mind that there is a small difference with other points of the domain if you want to use this property for something, but typically this is automatic since we will not move outside of the domain anyway. Nov 30, 2023 at 14:24

For points in the interior of the domain, you need to distinguish left from right continuity (i.e. depending on the side the limit is taken from). There, a function is continuous if it is both left and right continuous.

On the boundary of the function's domain, only the one-sided limits are permissible, since the limit approaching the boundary point from outside the domain will not be defined.

In the case of $$\sqrt x$$ at $$x_0=0$$, continuity can only be verified in terms of right continuity.

Generally speaking, a textbook about continuous real functions will proceed in this order:

• First, define continuity at a point contained in the domain of the function.
• Then, define left-continuity at a point contained in a left-open interval contained in the domain of the function, and define right-continuity at a point contained in a right-open interval contained in the domain of the function.
• Then, present a small theorem that says "If point $$x$$ is contained inside an open interval contained in the domain of function $$f$$, then $$f$$ is continuous at $$x$$ if and only if $$f$$ is both left-continuous and right-continuous at $$x$$."

In the case of $$f : [0, +\infty) \to \mathbb R\quad x \mapsto \sqrt x$$, this theorem doesn't apply at point 0 since 0 is not contained in an open interval contained in the domain of $$f$$; in fact it is not contained in a left-open interval contained in the domain of $$f$$, so there is no notion of left-continuity a this point.

Sometimes, textbooks will add the following small theorem, which applies in your case:

• If the domain of function $$f$$ is a left-closed interval $$[a, b)$$ or $$[a, b]$$, then continuity at $$a$$ is equivalent to right-continuity at $$a$$.

But sometimes textbooks will not bother to mention closed intervals at all, and will restrict their definitions to points which are contained in open intervals contained in the domain of the function.

Note that both of these theorems are an inevitable consequence of the definitions of continuity and right-continuity.

So, indeed, your function square root whose domain is $$[0, +\infty)$$ is both right-continuous and continuous at 0.