Answer
No. Please see below.
Work Step by Step
In the definition of “limit” of a function $f$ at a point $c$ we have “Let $f$ be a function defined on an open interval containing $c$ (except possibly at $c$).” But, there exists no interval containing $0$ such that the function $f(x)=\sqrt{x}$ is defined on it because the square root is not defined for any negative real number. Thus, $\lim_{x \to 0} \sqrt{x}$ does not exist.
