Prove that $\lim_{x \to 0} f(x) = \lim_{x \to 0}f(x^{3})$. Thought process behind steps in the solution. A question from Spivak's  Calculus asks me to prove that $$\lim_{x \to 0} f(x) = \lim_{x \to 0}f(x^{3})$$
Informally I understand what is going on, but trying to formalize it I ran into problems. Particularly there were two places where I struggled how to make the jump from one logical step to the next. After struggling I looked at a solution provided by Spivak which I am going to paste below:

The two issues I have is:
i) in the $\Rightarrow$ (I think this is the concluding "only if" direction). The step where Spivak states the assumption: 
$$0 < |x| < \text{min}(1, \delta)$$
How and why are we able to do this? I get that in all problems when looking for limits we are searching for a $\delta$, but since we are in essence unravelling definitions here why did we add in the assumption of "letting $\delta = 1$, doesn't that alter what we originally started with and restricted it more?
ii) in the opposite direction (concluding the "if" statement). There is the assumption that $$0 < |x| < \delta^{3}$$
Looking at it right now, I'm guessing the reason this is permitted is because we "have control" of $\delta$ and as such we can make it to be whatever we like. In this case since we want to work towards a specific result this $\delta$ works. 
Mechanically I see the reasoning as to why we want these things, but it is the abstract thinking about it that is still feeling weird to me. Perhaps I'm being too rigid when working with the definitions.
 A: Just for the sake of curiosity, I shall provide a more general result.
Let $g:X\to Y$ and $f:Y\to\textbf{R}$ such that $X\subseteq\textbf{R}$, $Y\subseteq\textbf{R}$, $x_{0}\in \textbf{R}$ is an adherent point of $E\subseteq X$ and $L\in\textbf{R}$ is an adherent point of $F\subseteq Y$. Thus, if $g\to L$ when $x\to x_{0}$ and $f\to M$ as $y\to L$, then we have that
\begin{align*}
\lim_{x\to x_{0};x\in E}f(g(x)) = M 
\end{align*}
Proof
Let $\varepsilon > 0$. Then there exists a $\delta_{1} > 0$ such that for every $y\in F$ satisfying $|y - L| < \delta_{1}$ we have that $|f(y) - M| < \varepsilon$.
On the other hand, for every $\delta_{1} > 0$, there is a $\delta > 0$ such that for every $x\in E$ satisfying $|x - x_{0}| < \delta$ we have that $|g(x) - L| < \delta_{1}$.
Since $g(x)\in Y$ for every $x\in X$, for every $\varepsilon > 0$ there corresponds a $\delta > 0$ such that for every $x\in E$ satisfying $|x - x_{0}| < \delta$ we have that $|f(g(x)) - M| < \varepsilon$, whence the desired result holds.
