Does $\lim_{x\to 0}\sqrt{x}$ exist?

Question

Since I'm an engineering student I have learned mathematics more in the intuitive way rather than formally. But I really like to understand mathematics formally and I'd like to take some courses with mathematicians. I've spent all my last semester trying to understand logic and set theory, and now I'm restudying again my course of calculus by myself. In this question I'm trying to understand the concept of limit. Then here is this thing that arose about the existence of $\lim_{x\to 0}\sqrt{x}$. In my course I learned that this doesn't exist because the funcion $f:(0,\infty)\longrightarrow \mathbb{R}$ such that $f(x)=\sqrt{x}$ is not defined on the the values before $0$, but now I'm not sure if that is the case or if this is so for engineering but for mathematicians this is different. Apparentely the limit does exist and is $0$ for mathematicians. I'd like to hear about it and all your comments about the definition of limit related to this.

Edit: This is the formal definition for limit that I have so far from the answer to my quesiton I said above (from which I can conclude that the limit of the expression I'm asking about is $0$):

Let $A, B\subseteq \mathbb{R}$ and $f:A\longrightarrow B$ a function such that $a\in \mathbb{R}$ is an acummulation point of $A$. Then we say that $l\in \mathbb{R}$ is the limit of the function $f$ when $x$ approches $a$ and is denoted by $\lim_{x\to a}f=l$ if and only if $\forall \epsilon\in \mathbb{R}(\epsilon>0)\exists\delta\in \mathbb{R}(\delta >0)\forall x\in A(0<|x-a|<\delta\longrightarrow |f(x)-l|<\epsilon)$.

Later LittleO told me that this is the definition found in Rudin's book of analyis: Definition 4.1

So, does $\lim_{x\to 0}\sqrt{x}$ exist or not? I'm really confused.

To be sure, you could denote this limit by $\lim_{x \to 0^+} \sqrt{x}$ (note the '+' sign) to make clear that you are approaching 0 from positive values of $x$ (i.e. where the function is defined). — tylerc0816, Sep 25 '13 at 00:04

score 1 · Accepted Answer · answered Sep 25 '13 at 01:01

1

It is an unfortunate necessity that any practical notation has to leave some things implicit to be filled in by the reader.

This situation is one of those. There are two very useful things we can say here:

Every interval around $0$ has points where $\sqrt{x}$ is not defined.
On the domain where $\sqrt{x}$ is defined, it approaches 0 as $x$ decreases to zero.

If the author does not explain which he means, you have to infer his intention from context.

answered Sep 25 '13 at 01:01

But then, does the limit exists or it doesn't? Are you suggesting that everything depends on the definition? – Daniela Diaz Sep 25 '13 at 01:06
mathematically THE limit x goes to zero only exists if the left limit agrees with the right sided limit. In your case it doesn't, because the left limit does not exist. So the limt does not exist. Make it into a right sided limit and then it exists and is zero. Check this out http://www.calculus-help.com/when-does-a-limit-exist/ – imranfat Sep 25 '13 at 02:24
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@Daniela: Yes-ish. There are at least two distinct concepts one might want to talk about, and in this particular situation, everything depends upon which of those concept is meant. And it's an unfortunate fact of life that many people won't realize that and muddy the waters by speak about what they consider to be the One True Meaning, and many people will realize that, but think which is meant is trivially obvious in any given situation, and will unfairly think you're being overly pedantic or splitting hairs when you try to express your confusion. – Sep 25 '13 at 02:24
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@Hurkyl, I'm not sure I agree. Analysis is very fiddly subject. A continuous function defined on a closed unit interval is bounded. If its defined on a half-open unit interval, it may be unbounded. A single point matters. So, I think that most definitions in analysis are either right or wrong. Right = the theory goes smoothly, wrong = its messy and full of caveats. – goblin GONE Sep 25 '13 at 03:00
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@user18921: Points only matter when they matter. If you're doing something where individual points don't matter, then you will cause problems by being picky about the points. The fundamental problem here (IMO) is the systematic confusion of functions and partial functions. Often, the distinction doesn't matter -- e.g. $x/x$ is not a function, but it's still obvious what we mean if we claimed it has limit at $0$, even though we usually only defined the limit for functions -- however the distinction matters for the OP's problem: there are at least two compelling candidates for the right thing. – Sep 25 '13 at 03:13
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Oh I agree about the partial functions. That's pretty much what I wrote here in my answer to the OP's original question. – goblin GONE Sep 25 '13 at 03:39

score 1 · Answer 2 · answered Sep 25 '13 at 04:42

To add another reference to the discussion, here is the definition of the statement $\lim_{x\to a} f(x) = b$ given in Calculus on Manifolds by Spivak:

In mathematical terms this means that for every number $\epsilon > 0$ there is a number $\delta > 0$ such that $|f(x) - b| < \epsilon$ for all $x$ in the domain of $f$ which satisfy $0 < |x - a| < \delta$. [emphasis is mine]

Unfortunately, Spivak uses a different definition in Calculus. In chapter 5 (p.104) he says

In order for $\lim_{x\to a} f(x)$ to be defined it is, as we know, not necessary for $f$ to be defined at $a$, nor is it necessary for $f$ to be defined at all points $x \neq a$. However, there must be some $\delta > 0$ such that $f(x)$ is defined for $x$ satisfying $0 < |x-a| < \delta$.

This statement contradicts the definitions given in baby Rudin and in Calculus on Manifolds. I think we should accept the definition in baby Rudin as standard.

score -1 · Answer 3 · answered Sep 25 '13 at 00:08

-1

Yes this limit must be zero in all senses, even if you define multivalued functions, i.e., a function assuming two or more values.

answered Sep 25 '13 at 00:08

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Does $\lim_{x\to 0}\sqrt{x}$ exist?

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