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.
 A: 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.
A: 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 $$|y-4|=\left|(\sqrt{y})^2-2^2\right|=|(\sqrt{y}+2)(\sqrt{y}-2)|=|\sqrt{y}+2|\cdot|\sqrt{y}-2|\tag{2}$$ wherever each expression is defined. Thus, noting that $\sqrt{y}+2$ is at least $2$ for any 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.
A: We have a function $f(x) = \sqrt{4-x}$
$\tag 1 f: (-\infty,4] \to [0, +\infty)$
and the OP is encouraged to plot it before working on the problem.
To solve the following two exercises apply algebraic and analytic techniques to the quadratic inequalities that result from 'squaring both sides'.
Exercise 1: Show that
$\quad f(x) \lt 2 -x$ if and only if $x \lt 0$.
Hint: Check for solutions on the subintervals $(-\infty,0]$, $(0, 2]$ and $(2, 4]$.
Exercise 2: Show that
$\quad f(x) \gt  2 -\frac{x}{2}$ if and only if $0 \lt x \lt 4$.
Hint: Check for solutions on the subinterval $(-\infty,4]$.
Using the above and further arguments we can write as true the following two statements: 
$\tag 2 x \lt 0 \; \implies 2 \lt  f(x) \lt 2 -x$
$\tag 3 x \in (0,4) \; \implies 2 -\frac{x}{2} \lt  f(x) \lt 2$
Now using the squeeze theorem we have
$\tag 4 \displaystyle \mathop {\lim }\limits_{x \to 0^{-}} \sqrt {4 - x} = 2$
$\tag 5 \displaystyle \mathop {\lim }\limits_{x \to 0^{+}} \sqrt {4 - x} = 2$
and so indeed (see this),
$\tag 6 \displaystyle \mathop {\lim }\limits_{x \to 0} \sqrt {4 - x} = 2$
Exercise 3: Use the above to 'back-out' an Epsilon/Delta argument showing that $\text{(6)}$ is true.
