Based on your correction, I recommend the following approach.
If we set $y:=4-x$ (to clean up the expression), then $x\to 0$ precisely as $y\to 4$, yes? Thus, we may equivalently show that $\lim_{y\to 4}\sqrt{y}=2.\tag{1}$ Fixing some $\varepsilon>0$, we must find $\delta>0$ such that for $0<|y-4|<\delta$, we have $|\sqrt{y}-2|<\varepsilon$. It's worth noting here that $|y-4|=|(\sqrt{y}+2)(\sqrt{y}-2)|=|\sqrt{y}+2|\cdot|\sqrt{y}-2|\tag{2}$ wherever each expression is defined. Thus, noting that $\sqrt{y}+2$ is positive for positive real $y$, we have $\frac1{\sqrt{y}+2}\leq\frac12<1\tag{3}$ for positive real $y$. Hence, for any $\delta>0$, we have by $(2)$ and $(3)$ that $0<|y-4|<\delta\quad\Rightarrow\quad|\sqrt{y}-2|<\frac1{\sqrt{y}+2}\delta<\delta,\tag{4}$ wherever each expression is defined.
Now, it looks like we can just set $\delta=\varepsilon$ and be done with it, yes? There's only one potential problem: What if one of the expressions is undefined? Indeed, if our arbitrary $\varepsilon>0$ is too large, we may well be making claims involving square roots of negative numbers--for example, if we had chosen $\varepsilon=5$, then $y=-1/2$ satisfies $|y-4|<\varepsilon$, but $|\sqrt{y}-2|$ doesn't make any sense (in this context). Thus, to ensure that $y\geq 0$, we need $|y-4|\leq 4$, so we set $\delta:=\min\{4,\varepsilon\},$ whence each expression in $(4)$ is defined, and $0<|y-4|<\delta\quad\Rightarrow\quad|\sqrt{y}-2|<\delta\leq\varepsilon,\tag{5}$ as desired.
To get the result in terms of $x$, we need only substitute $y=4-x$ in each instance, and observe that $|y-4|=|-x|=|x|=|x-0|$, so everything looks just as it should, and shows that the two approaches (using $x$ vs. using $y=4-x$) are indeed equivalent.