A limit exists iff and only the left limit and the right limit exist and are equal to each other It is well known that

$$\lim_{x \to a} f(x) = L \iff \lim_{x \to a+}f(x) = L = \lim_{x \to
 a-}f(x)$$

Consider the function $\sqrt{.}: \mathbb{R}^+ \to \mathbb{R}$
Now, consider $\lim_{x \to 0} \sqrt{x}$
We can prove this limit is equal to $0$. Indeed, let $\epsilon > 0$. Choose $\delta = \epsilon^2$. Then, for $x \in \mathbb{R}^+$ satysfying $0 < |x| < \delta$ or equivalently $0 < x < \delta$, we have $\sqrt{x} < \sqrt{\delta} = \epsilon$, which establishes the result.
However, my confusion lies in the following: the limit from the left does not seem to exist, making the above theorem untrue. Where lies my mistake?
 A: Actually, what you wrote is not well known, and is in fact not even true.
What is well known is the fact that

Let $f$ be a function $\color{red}{\text{defined on some open neighborhood of $a$}}$. Then, $$\lim_{x \to a} f(x) = L \iff \lim_{x \to a+}f(x) = L = \lim_{x \to
 a-}f(x)$$

Always read the fine print.
A: In fact the limit exists in $x_0$ if and only if for all $x$ in the domain of the function we have$$\forall \epsilon>0\qquad,\qquad \exists \delta\qquad |x-x_0|<\delta\to |f(x)-f(x_0)|<\epsilon$$and the domain of $f(x)=\sqrt x$ in $x_0=0$ is only $\{x\ge 0\}$ where only the right limit needs to exist. This also can be generalized further to higher dimensions where all limits from all valid directions in a sufficiently small neighborhood of some point over a metric space of metric $d(.,.)$ must exist and be equal i.e.$$\forall \epsilon>0\qquad,\qquad \exists \delta\qquad d(x,x_0)<\delta\to d(f(x),f(x_0))<\epsilon$$
A: You should be careful while  applying the $\varepsilon-\delta  $ definition it says. Let $a$ be an accumulation (or limit point ) of $Dom(f)$ then $f$ is continuous at  $a$ if  
$$\forall~~\varepsilon>0, \exists~\delta>0:\color{blue}{\text{for every $x\in Dom(f),$  } 0<|x-a|<\delta }\implies |f(x)-f(a)|<\varepsilon $$
in your case $x<0$ is not in the domain of $\sqrt{\cdot}$
And then the notion of limit from the left at $x=0$ does not apply here, 
