Does $\lim_{x\to 0}\sqrt{x}$ exist? 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.
 A: 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.
A: 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.
A: Yes this limit must be zero in all senses, even if you define multivalued functions, i.e., a function assuming two or more values.
