# Is $\lim_{x\rightarrow 0}\sqrt x$ well-defined? [duplicate]

This question already has an answer here:

I have a limit, $$\sqrt x$$, in which I consider only values of $$x\in\mathbb{R}$$. If I then consider values of $$x$$ approaching zero, I know that:

$$\lim_{x\rightarrow 0^{-}}\sqrt x$$

does not exist, and that:

$$\lim_{x\rightarrow 0^{+}}\sqrt x$$

does exist, so I can then conclude that the limit does not exist and I understand that, but does this mean that the limit is not well defined as $$x$$ approaches $$0$$ as well?

Thanks.

## marked as duplicate by Hans Lundmark, YuiTo Cheng, Xander Henderson, The Count, воительJul 15 at 1:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• – Monadologie Jul 14 at 20:04

## 2 Answers

If $$f(x)=\sqrt{x}$$, we consider $$f(0)$$ something called an endpoint discontinuity. How can it be discontinuous, though? Remember the limit definition of continuity:

$$f(0)$$ is continuous iff: $$\lim_{x\to 0^-} \sqrt{x} = \lim_{x\to 0^+} \sqrt{x}$$ $$\lim_{x \to 0}=f(0)$$

$$f(0)$$ fails the first part of the test and the limit does not exist because you can't approach it from the left. If you want, you could prove this with the formal definition of a limit, but we could also use intuition. In your question, you said we are only looking at values of $$x \in \mathbb R$$, therefore we can have no value less than $$0$$ (for example, $$\sqrt{-1}$$) and the two-sided limit does not exist, therefore an endpoint discontinuity exists at $$f(0)$$.

• I understand that the limit does not exist, but to clarify I wanted to know that, in this instance, even though the limit does not exist, is it still considered well defined at 0? Or does a limit that is well defined need to exist to be considered well defined? – biscuitflipper Jul 15 at 3:20
• Well, the usual concept of a limit doesn't properly apply to an endpoint. You should look at some of the answers in math.stackexchange.com/questions/637280/… – N. Bar Jul 15 at 3:24

For the function $$f(x)=\sqrt{x}$$, unless specified otherwise, we regard it as a function from $$[0,\infty)$$ to $$\mathbb{R}$$.

Since the domain is $$[0,\infty)$$, we can't approach $$0$$ from the left, hence in this context, since $$0$$ is a left endpoint of the domain, the phrase "$$x$$ approaches $$0$$" means "$$x$$ approaches $$0$$ from the right".

Hence we have $$\lim_{x\to 0}\sqrt{x}=\lim_{x\to 0^{+}}\sqrt{x}=0$$.

• So given that the two-sided limit does not exist, does this also mean that the two-sided limit is not well defined here? Does the existence of both right and left hand sided limits make the limit at a point well defined? – biscuitflipper Jul 14 at 20:27
• @biscuitflipper: If the domain of a function has an endpoint, say $a$, there's no concept of two-sided limit at $x=a$. But in that case, the (ordinary) limit as $x$ approaches $a$ defaults to the appropriate one-sided limit. The concept of limit assumes the approach is from within the domain. In particular, at an endpoint I wouldn't say that the two-sided limit does not exist, since that suggests that both one-sided limits were analyzed. Instead I would say that an endpoint, analyzing a limit means analyzing the appropriate one-sided limit. – quasi Jul 14 at 20:40