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.
You first need a suitable candidate/educated guess for what the limit ought to be. Then, only after that, you can use the precise definition to PROVE that your initial guess is indeed the case. Also, you can see that this is the best you can do simply from how the definition of limits is given:
Definition.
Let $f:\Bbb{R}\to \Bbb{R}$ be a function, $a\in\Bbb{R}$. We say $f$ has a finite limit at $a$ if there exists $l\in \Bbb{R}$ such that for every $\epsilon>0$, there exists $\delta>0$ such that for all $x\in\Bbb{R}$, if $0<|x-a|<\delta$ then $|f(x)-l|< \epsilon$.
(In this case, we can prove that $l$ is unique and we denote it as $\lim_{x\to a}f(x)$)
Notice how the definition starts with "there exists $l\in \Bbb{R} \dots$" Just from the way it is phrased, it suggests that before even checking the $\epsilon,\delta$ criterion, you need to have a candidate value for the limit $l$. Nowhere does the definition tell you what $l$ is or how to go about guessing this (such "guess work" is something you pick up along the way as you learn more).
For example, if you had two functions $f$ and $g$, with $\lim\limits_{x\to a}f(x) = l_1$ and $\lim\limits_{x\to a}g(x) = l_2$, then if all you do is stare at the definition of limits, there's no way you can tell that $f+g$ also has a limit and that the limit equals $l_1+l_2$. The only natural guess would be that if $f+g$ had a limit, then it had better be $l_1+l_2$.
Then, once you have this guess, you then proceed to prove this using the precise $\epsilon,\delta$ definition (where the crux of the proof is the triangle inequality).
Best Answer
Note that\begin{align}\frac{x^3-1}{x-1}-3&=x^2+x+1-3\\&=x^2+x-2\\&=x^2-1+x-1\\&=(x-1)(x+1)+x-1.\end{align}Now, take $\varepsilon>0$. If $\lvert x-1\rvert<1$, then $\lvert x+1\rvert<3$ and therefore$$\bigl\lvert(x-1)(x+1)+x-1\bigr\rvert<4\lvert x-1\rvert.$$So, take $\delta=\min\left\{1,\frac\varepsilon4\right\}$ and then$$\lvert x-1\rvert<\delta\implies\bigl\lvert(x-1)(x+1)+x-1\bigr\rvert<4\times\frac\varepsilon4=\varepsilon.$$