Prove $\lim_{x\to0}\sqrt{4-x}=2$ using the precise definition of limits. (Epsilon-Delta) Prove $$\lim_{x\to0}\sqrt{4-x}=2$$ using the precise definition of limits. (Epsilon-Delta)
I am not sure how to link $0<\left |x  \right |<\delta$ with $\left |\sqrt{4-x}-2\right |<\epsilon$ .
EDIT (Trying it out now)
I worked till here, then I basically got stuck.

First, you can try and prove some important yet simple facts about limits. The following are all equivalent:
$$\eqalign{
  & \mathop {\lim }\limits_{x \to a} f\left( x \right) = l  \cr 
  & \mathop {\lim }\limits_{h \to 0} f\left( {a + h} \right) = l  \cr 
 & \mathop {\lim }\limits_{h \to 0} f\left( {a- h} \right) = l \cr
  & \mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - l} \right] = 0 \cr} $$
You have 
$$\mathop {\lim }\limits_{x \to 0} \sqrt {4 - x}  + 2 = 4$$
or
$$\mathop {\lim }\limits_{h \to 0} \sqrt {4 - h}  + 2 = 4$$
By the above, this is equivalent to
$$\mathop {\lim }\limits_{x \to 4} \sqrt x  + 2 = 4$$
So you want to show that for each $\epsilon >0$ there is a $\delta>0$ such that for all $x$, $$0 < \left| {x - 4} \right| < \delta  \Rightarrow \left| {\sqrt x  - 2} \right| < \varepsilon $$
But multiplying the conjugate gives $$\left| {\sqrt x  - 2} \right| = \left| {\frac{{x - 4}}{{\sqrt x  + 2}}} \right| < \left| {x - 4} \right|$$
so taking $\delta=\epsilon$ does it.
NOTE You could've also worked with $$\mathop {\lim }\limits_{x \to 0} \sqrt {4 - x}  + 2 = 4$$
in fact,
$$\left| {\sqrt {4 - x}  + 2 - 4} \right| = \left| {\sqrt {4 - x}  - 2} \right| = \left| {\frac{{ - x}}{{\sqrt {4 - x}  + 2}}} \right| < \left| x \right|$$
so again, $\delta=\epsilon$, as expected.
