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?
 A: I recall that a function $f:\Bbb R\to \Bbb R$ is continuous at a point $x$
provided for each $\varepsilon>0$ there exists $\delta>0$ such that for each point $x’$ with $|x’-x|<\delta$ holds $|f(x’)-f(x)|<\varepsilon$.
When we apply this definition to functions defined only on a proper subset $X$ of $\Bbb R$ (for instance, a function $\sqrt{x}$ is defined on a set $\Bbb R_+=\{x\in\Bbb R: x\ge 0\}$) we can encounter a problem if it will happen that for some point $x\in X$ for each $\delta>0$ there exists a point $x’\in\Bbb R$ such that $|x’-x|<\delta$, but $x’$ lies outside $X$ and therefore the value of $f(x’)$ is undefined (as $\sqrt{x’}$ is undefined for negative $x’$). 
This problem is solved by modifying the continuity definition by adding to it condition that $x’$ should belong to the domain of $f$, that is if $X$ is a subset of $\Bbb R$ then a function $f:X\to \Bbb R$ is continuous at a point $x\in X$ provided for each $\varepsilon>0$ there exists $\delta>0$ such that for each point $x’\in X$ with $|x’-x|<\delta$ holds $|f(x’)-f(x)|<\varepsilon$. 
This modification is formally (that is set-theoretically, or really or true, if you wish) correct because when we say that $f$ is a function from a set $X$ to a set $Y$ we mean that for each element $f(x)$ of $X$ there is a unique element $f(x)$ of $Y$. As use set, the domain of the function is included in its definition. Also this modification corresponds to common and very general (that is real) definitions of continuity (see the appendix).
But in calculus when we consider functions defined only on proper subsets of $\Bbb R$, for instance, rays or segments, can be useful notions of one-sided continuity. These are not a continuity in strict sence, but became it when we restrict the domain of the function to the corresponding side. 
For instance, the formal answer to your question is the following. A function $\sqrt{x}$ from $\Bbb R$ to $\Bbb R$ is not continuous at $0$, it is not even a function on $\Bbb R$. But a function $\sqrt{x}$ from $[0,\infty)$ to $\Bbb R$ is continuous at $0$.
I also note that when a function is continuous both from the left and from the right then its one-side continuities may be thought as kinds of half-continiuty, :-) which union with the other half coincides with the true continuity. 
Appendix.  General definitions of continuity.
We can straightforwardly generalize the $\varepsilon$-$\delta$-continuity defintion from $\Bbb R$ to so-called metric spaces. A function $f$ from a metric space $(X,d_X)$ to a metric space $(Y,d_Y)$ is continuous at a point $x$ provided for each $\varepsilon>0$ there exists $\delta>0$ such that for each point $x’\in X$ with $d_X(x’,x)<\delta$ holds $d_Y(f(x’),f(x))<\varepsilon$. By the way, it is equivalent to the following definition: if a sequence $\{x_n\}$ (of points of $X$) converges to $x$ then a sequence $\{f(x_n)\}$ (of points of $Y$) converges to $\{f(x)\}$. Moreover, we can generalize continuity from metric spaces to much more general topological spaces.
A function $f$ from a topological space $(X,\tau_X)$ to a topological space $(Y,\tau_Y)$ is continuous at a point $x$ provided for each neighborhood $O_{f(x)}\in\tau_Y$ of the point $f(x)$ there exists a neighborhood $O_x\in\tau_X$ of the point $f(x)$ such that $f(O_x)\subset O_{f(x)}$. 
Remark, that all these definitions require that the function $f$ is defined on the whole space $X$.
A: The distinction between a 'left-side' limit, 'right-side' limit and (a general) limit makes sense e.g. for piece-wise functions at endpoints of pieces.
In a case like this one, when the domain is an interval, there is no need to specify wheteher we consider the limit or continuity at the left endpoint of the interval from the right, because in such case there exists only the limit and continuity from the right. 
A: It is correct to say that $\lim_{x \to 0}\sqrt{x}=0$ without specifiying, but it's incorrect to say that $f$ is continuous at $x=0$. It is only continuous from the right.
Furthermore, a function is continuous at a point if and only if the left and right hand limits are equal. Since this isn't the case $f$ can't be continuous at $x=0$ in the general sense.
A: 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 $a=0$.
A: 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. 
A: It all depends on where the function is defined. In your example, the function is not defined for negative numbers. And at zero, the value of function is zero.
 And since you can approach zero, from the direction where function is defined, that is from right, the value goes tends to zero. Hence the function is continuous. Meaning continuous on all points where it is defined. But at zero, left continuity doesn't exist at zero. So according to the real definition of continuity, it is not continuous at zero
