Square root limit with $\epsilon-\delta$
Since $\sqrt x$ is undefined when $x<0$, it makes no sense to talk about $\lim_{x\to0^-}\sqrt x$. And it also follows that $\lim_{x\to0^+}\sqrt x=\lim_{x\to0}\sqrt x$, since both assertions $\lim_{x\to0^+}\sqrt x=0$ and $\lim_{x\to0}\sqrt x=0$ mean$$(\forall\varepsilon>0)(\exists\delta>0):|x|<\delta\wedge x>0\implies\left|\sqrt x\right|<\varepsilon.$$