$\lim_{x\to a^{-}} f(x) =\lim_{x\to a^{+}} f(x) =L$ We know that :
$$\lim_{x\to a} f(x) =L$$
if and only if :
$$\lim_{x\to a^{-}} f(x) =\lim_{x\to a^{+}} f(x) =L$$
now :
let $$f(x)=\sqrt{x}\\\lim_{x\to 0^{+}} \sqrt{x} =0\\\lim_{x\to 0^{-}} \sqrt{x} =!!!$$
so :
$$\lim_{x\to 0} \sqrt{x}= \text{Does not exist}$$
is it right ?
 A: When you give the formal definition of the limit you have to be careful to where you are define things. Suppose $A\subseteq \mathbb R$ and $f: A \to \mathbb R$. Next, pick $a \in \mathbb R$. The symbols you use
$$\lim_{x \to a^+} f(x), \quad \lim_{x \to a_-} f(x)$$
make sense only if the sets $A \cap (a,a+\delta)$ and $A \cap (a-\delta,a)$  are not empty for all $\delta >0$. This is not the case with $f(x) = \sqrt{x}$ and $a =0$, as $(0,+\infty) \cap (-\delta,0)$ is empty for all $\delta$. In this case we say that $f$ has limit at $0$ provided only
$$\lim_{x \to 0^+} f(x)$$
exists.
A: No, the conclusion is not right. The problem is here to use a statement which is normally true and apply it to a situation where it's not applicable/true.
The statement that $\lim_{x\to a}f(x) = L$ if and only if $\lim_{x\to a^{-}} f(x) =\lim_{x\to a^{+}} f(x) =L$ is only true under certain circumstances. It requires $f$ to be defined sufficiently close to $a$ at both sides, more precisely it means that for every $\delta>0$, $f$ must be defined for some $a-\delta<x<a$ and $a<x<a+\delta$.
This means that the statement is not valid for $f(x)=\sqrt x$ and $a=0$ since $f(x)$ is not defined for any $x<0$, it means for any given $\delta >0$ there is no $-\delta<x<0$ such that $f(x)$ is defined.
The statement can be altered to handle other cases. If for example $f(x)$ is never defined for $x<a$ and for each $\delta>0$ there exists an $x$ with $a<x<a+\delta$ for which $f(x)$ is defined then $\lim_{x\to a} f(x)=L$ if and only if $\lim_{x\to a^+}f(x)=L$
A: This 'problem' is avoided by restricting the $x$-values around $a$ you take into consideration to values in the domain of $f$, since $f(x)$ does not make any sense if $x$ is not in the domain.
The equivalence between the existence of a limit at $a$ and the existence of both one-sides limits at $a$, is only valid if $f$ is defined on the left and on the right of $a$.
