Question on the Validity of $\lim_{x\to a} {x^3} = {a^3}$ Proof from Spivak's Calculus I was reading through the limits chapter, when I stumbled upon the proof that $f(x) = x^3$ approaches $a^3$ as $x$ approaches $a$

If $$|x - a| < \min \left(1, \frac{\epsilon}{(1+|a|)^2 + |a|(1+|a|) + |a|^2} \right)$$ then $|x^3 - a^3|$ $ < \epsilon$.

This is what Spivak did

If $|(x-a)| < 1$, then $|x| < |a| + 1$ and consequently $$|x^2 + ax +
 a^2| \leq |x|^2 + |a||x| + |a|^2 < (1+|a|)^2 + |a|(1+|a|) + |a|^2 $$
Therefore  \begin{align} |x^3 - a^3| &= |x-a||x^2 + ax + a^2| \\ &<
 \frac{\epsilon}{(1+|a|)^2 + |a|(1+|a|) + |a|^2}(1+|a|)^2 +
 |a|(1+|a|) + |a|^2 \\ &= \epsilon \end{align}

I tried doing it on my own before seeing the result and I got
\begin{align}
|x^2 + ax + a^2| &\leq |x|^2 + |a||x| + |a|^2 \\
&< (2|a| + 1)^2 \\
&= 4a^2 + 4|a| + 1
\end{align}
which is almost the same since $(1+|a|)^2 + |a|(1+|a|) + |a|^2 = 3a^2 + 3|a| + 1$.
So I was wondering if my answer is invalid or not, and if not, does that mean some limits have several possible $\delta$ values when $x$ approaches some $a$? 

Once again, it doesn't matter that you arrive at $\epsilon$ precisely!
If for $|x-a|<\delta$ you can show that $|x^3-a^3|<K\epsilon$ with $K$ a constant (i.e. independent of $x$) then it's a win.
Cutting epsilons is just a matter of aesthetic and IMHO just confuse the beginner.
This practice is a bit outdated (we nowadays make use of equivalents much more than in the past), and it is not mandatory to have to go for the finest inequalities (i.e. $K=1$) in delta-epsilon proofs, you are just as fine with $k=6446595379f(a)$ or any other value whatsoever...
So you get $|x-a|<\delta$.
For the convergence part you need $\delta<\epsilon$
For the bounded part, you need $|x|<|a|+\delta$ so just do it VERY roughly and choose $\delta<|a|$ (why? well, because it simplifies all calculations).
Then you get $|x|<2|a|$ and $|x^2+ax+a^2|<4a^2+2a^2+a^2=7a^2$
$K=7a^2$ is constant, that is all you need.
Conclude by taking $\delta=\min(\epsilon,|a|)\implies |x^3-a^3|<7a^2\epsilon$.
