Limit of functions on a undefined neighborhood I am reading a textbook about Real Analysis by Trench.
He defines the limit $L$ of a function $f(x)$ as $x$ approaches $x_0$ as follows (p. 34):
$$\lim_{x \rightarrow x_0} f(x) = L$$
if $f$ is defined on some deleted neighborhood of $x_0$, and for every $\epsilon > 0$, there is a $\delta > 0$ such that
$$|x - x_0| < \delta \Rightarrow |f(x) - L| < \epsilon$$ 
That's fine so far. But then he continues and says that for the function $f(x) = 2x \sin\sqrt{x}$ the limit $\lim_{x \rightarrow 0} f(x) \neq 0$ because $f$ is not defined for negative x (p. 37 - 38).
He then says later on that for the function $f(x) = \sqrt{x}$ the limit $\lim_{x \rightarrow x_0} = f(x_0)$, where $x_0$ can also be $0$ (p. 56). How does that work? What did I miss? The second functions isn't defined for negative $x$ as well, just as the first function wasn't. Why does it work this time for $x_0 = 0$ and why didn't it work for the first function?
Edit: Well technically he didn't say that $x_0 = 0$ in his proof, but he still claims at the end that it does work for $x_0 = 0$, even though he has only shown that $\lim_{x \rightarrow 0^+} = 0$ 
 A: Accoding to the def'n of limit quoted (p. 34) the limit of the real-valued function $f(x)=2x \sin \sqrt x\;$ as $x\to 0$ does not exist because the def'n requires that $f(x)$ exists  for every $x \in (-r,0)\cup (0,r)$ for some $r>0.$ I get the impression the author is careless here, or later. If $f$ is defined on $D\subset  \Bbb R$ then we usually write $\lim_{x\to x_0}f(x)=L$ to mean that $x_0\in \overline D$ and that $$\forall e>0\;\exists d>0\;\forall x\in D\cap (-d+x_0,d+x_0)\;(|f(x)-L|<e).$$ 
We  require that $x_0\in \overline D$ because the above line, by itself, is logically true for  $any$   $L$ if $x\not\in \overline D$ (Because if $d>0$ is small enough that $D\cap (-d+x_0,d+x_0)$ then there is no $x\in D\cap (-d+x_0,d+x_0)$ for which the condition $|f(x)-L|<e$ is false). 
Note that this def'n applies also when $x_0\in D.$ When $x_0\in D$ we are often concerned with the limit of $f(x)$ as $x$ approaches $x_0$ thru members of $D$ other than $x_0$ itself. There does not seem to be a "one-word" symbol for this.    
A: $lim_{x\to0}\sqrt{x}$ doesn't exist because $lim_{x\to0^-}\sqrt{x}$ doesn't exist, if the text says the opposite then it's maybe talking in other sense.
