TITLE: $\epsilon$-$\delta$ limits of functions question
QUESTION [8 upvotes]: I'm studying how to write epsilon-delta proofs for limits of sequences, limits of functions, continuity, and differentiability and I'm having trouble with the general methodological procedure used in some of the proofs in the text as opposed to some of the proofs I have come up with. I'm going to post an example question with the proof I came up with and the proof given by the book. If possible, could you comment on how my proof is insufficient to prove the result, because I don't understand why it isn't.
I understand the definition of the limit of a function to be the following:

$\lim_{x \rightarrow a} f(x) = L$ means, $\forall \epsilon >0, \exists \delta >0; 0 < |x-a|< \delta \Rightarrow |f(x) - L|< \epsilon$, with $x, a, L \in \mathbb{R}$.

Question:
Give an $\epsilon$-$\delta$ proof that $\lim_{x \rightarrow 2} x^3 =8$.
My Proof:

Let $\epsilon > 0$ be arbitrary and for $\delta = (\epsilon +8)^\frac{1}{3} -2$ assume that $ 0 < |x-2| < \delta$, with $x \in \mathbb{R}$. Then, 
  $$|x-2| < (\epsilon +8)^\frac{1}{3} -2$$
  $$x-2 < (\epsilon + 8)^\frac{1}{3} -2 $$
  $$x< (\epsilon +8)^\frac{1}{3}$$
  $$x^3 < \epsilon +8$$
  $$x^3 - 8 < \epsilon$$
  $$|x^3 - 8| < \epsilon$$
  Therefore, for $f(x) = x^3$ and $L = 8$, we have shown $|f(x) -L| < \epsilon$ and resultantly $\lim_{n \rightarrow 2} x^3 = 8$.

Book's Proof:

Let $\epsilon >0$ be given and choose $\delta = \min(1, \frac{\epsilon}{19})$. Let $x \in \mathbb{R}$ such that $0 < |x-2| < \delta = \min(1,\frac{\epsilon}{19})$. Since $|x-2|<1$, it follows that $ -1 < x-2 < 1$ and so $1 <x<3$. Thus $|x^2 + 2x +4| < 19$. Because $|x-2| < \frac{\epsilon}{19}$, it follows that $|x^3 - 8| = |x-2||x^2 + 2x +4| < 19|x-2|< 19(\frac{\epsilon}{19})=\epsilon$.

Aren't we trying to show that given our assumption that $0 < |x-a|< \delta$ we need to show that $|f(x) - L|< \epsilon$. That is all that is required right? We pick some $\delta$ for which this works? I'm able to follow to proof in the book and understand that it is correct, I just don't understand why all of the additional details are necessary.
Thank you for any responses.

REPLY [3 votes]: Here is an alternative approach to this problem that builds up some background instead of attacking it directly. The value in this is that you can solve all problems of the given sort instantly, after having set up the machinery.

I define $F$ to be the set of functions such that if $f \in F$ then $\lim_{x \rightarrow a} f(x) = f(a)$. Given that we can prove the theorem in the following way:

$2^3 = 8$ proves that $\lim_{x \rightarrow 2} x^3 = 8$ since $x^3 \in F$.

To see that $x^3 \in F$ I can prove that $x \in F$ and that if $f, g \in F$ then $fg \in F$.

Theorem $x \in F$. This is equivalent to $\lim_{x \to a} x = a$ which is, by definition, $$\forall \varepsilon > 0, \exists \delta > 0, \forall x, 0 < |x - a| < \delta \implies |x - a| < \varepsilon.$$ This is proved by letting $\delta = \epsilon$.
Theorem $f,g \in F \implies fg \in F.$ This is equivalent to saying that if $\lim_{x \to a} f(x) = f(a)$ and $\lim_{x \to a} g(x) = g(a)$ then $\lim_{x \to a} f(x)g(x) = f(a)g(a).$ To prove this we will use the definitions of our two hypotheses:
$$\forall \varepsilon_1 > 0, \exists \delta_1 > 0, \forall x, 0 < |x - a| < \delta_1 \implies |f(x) - f(a)| < \varepsilon_1$$
$$\forall \varepsilon_2 > 0, \exists \delta_2 > 0, \forall x, 0 < |x - a| < \delta_2 \implies |g(x) - g(a)| < \varepsilon_2$$
to prove that $$\forall \varepsilon > 0, \exists \delta > 0, \forall x, 0 < |x - a| < \delta \implies |f(x)g(x) - f(a)g(a)| < \varepsilon.$$
To prove this part you can use the triangle inequality (specifically $|f(x)g(x) - f(a)g(a)| \le |f(x)g(x) - f(x)g(a)| + |f(x)g(a) - f(a)g(a)|$) and find a suitable $\delta$ in terms of $\varepsilon_1$ and $\varepsilon_2$. I leave that as an exercise.

Exercise Prove that constants and sums are in $F$, use it to prove that $\lim_{x \rightarrow 2} 1 - x^3 + (7-3 x-2 x^2)^4 = 2394$ in one line.