Continuity and limits at end point of interval I am bad a calculus and I have question about continuity. If I have a polynomial, then the function is continuous on $\mathbb{R}$ because $\lim_{x\to a} f(x) = f(a)$ for all $a\in \mathbb{R}$.
My question is if $f(x) = \sqrt{x}$ is continuous at $0$. My text book doesn't say if this is continuous at $0$. It does say that the function is continuous from the right. But I have also heard that continuity is a more general concept.
My question is: under the real definition of continuity, if $f(x) = \sqrt{x}$ continuous at $0$?
Is it correct to say that $\lim_{x\to 0} \sqrt{x} = 0$ without specifying that $x$ is approaching from the right?

For $f(x)=\sqrt{x}$ to be continuous at $a$, $\lim_{x \rightarrow a}f(x)=f(a)$. Let's subsitute $0$ for $a$, and now we know we have to prove that $\lim_{x \rightarrow 0}f(x)=f(0)=0$. For this to be true, we have to have two things. 
$$\lim_{x \rightarrow 0+}f(x)=0$$
$$\lim_{x \rightarrow 0-}f(x)=0$$
Firstly, it's clear to see that the first inequality holds true. 
However, the second inequality is where things get tricky. According to the definition of a limit (which is one of the biggest confusing jumble of symbols you'll encounter in Calc 1), we need to be able to evaluate the function from points less than $0$ to "approach" $0$ from the left, which is what we are doing in the second inequality. However, for numbers less than $0$, $\sqrt x$ is not defined when $f(x) < 0$, so we have no way to evaluate the second inequality. 
Therefore, $f(x)$ is not continuous at $0$ because $\lim_{x \rightarrow 0}f(x) \neq 0$, and that's because $\lim_{x\rightarrow0-} \neq 0$ - it doesn't equal anything, in fact.  
Side note: If $x$ was taken to be a complex number, which I assume it is not from your definition, $\sqrt x$  (graphed as a vector field) would be continuous at 0. Someone correct me if I'm wrong. 
