Answer
See explanations.
Work Step by Step
Step 1. Let $f(x)=\sqrt {4-x}, c=0, L=2$; the limit becomes $\lim_{x\to c}f(x)=L$
Step 2. For a given small value $\epsilon\gt0$, we need to find a value $\delta\gt0$ so that for all $x$ when $0\lt |x-c|\lt \delta$, we have $|f(x)-L|\lt\epsilon$
Step 3. The last inequality from the last step gives $|\sqrt {4-x}-2|\lt\epsilon$ which gives $-\epsilon\lt \sqrt {4-x}-2\lt\epsilon$ or $2-\epsilon\lt \sqrt {4-x}\lt2+\epsilon$
Step 4. Assume $x\leq4$; we can get $(2-\epsilon)^2\lt4-x \lt(2+\epsilon)^2$ which leads to $4-(2+\epsilon)^2\lt x\lt4-(2-\epsilon)^2$
Step 5. Compare the result from step 4 with the requirement for the value $\delta$ in step 2; we can set $\delta$ as the smaller value of $|4-(2+\epsilon)^2|$ and $|4-(2-\epsilon)^2|$
Step 6. With the setting of the $\delta$ value, we can go backwards through the steps of 4,3,2 to reach a conclusion that for all $x$ when $0\lt |x|\lt \delta$, we have $|\sqrt {4-x}-2|\lt\epsilon$ which proves the limit statement.
