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?

We say a function $f$ is continuous at some interior point $a$ $\iff$
$$\forall \epsilon >0, \exists \ \delta>0, \ s.t. \ |x-a|<\delta\implies|f(x)-f(a)|<\epsilon$$
Or more simply notated as:
$$\lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)=f(a)$$
Considering one sided limits, we say $\lim{_{x\to a^+}f(x)}$ exists and equals $L \iff$
$$\forall\epsilon > 0, \exists \ \delta > 0 \ s.t. \ 0<x-a<\delta \implies|f(x)-L|<\epsilon$$
Such that $L$ is the limit at some exterior point $a$.
We may then specify this further to the case of endpoints, and define the following:
A function is continuous from the right at some exterior point $a$ if $\lim_{x\to\ a^+}f(x)=f(a)$, and is continuous from the left  if $\lim_{x\to\ a^-}f(x)=f(a)$
Thus we may say that $f(x)=\sqrt{x}$ is continuous from the right at the point $a=0$.
