Spivak Calculus Chapter 5 (Limits) Problem 3 Part (i) The question is: Find a $\delta$ such that $|f(x)-l|<\epsilon$ for all $x$ satisfying 0 < $|x-a|$<$\delta$. 
(i) $f(x)=x^4; l=a^4$
This was my attempt, but I am lacking confidence in every single attempt of Spivak's exercises because I feel I not as rigorous as Spivak wants me to be.:
$|x^4-a^4|<\epsilon\\|x^4|-|a^4|<\epsilon\\|x|-|a|<\frac{\epsilon}{(|x|+|a|)\cdot (|x|^2+|a|^2)}$
So, $\delta=\frac{\epsilon}{(|x|+|a|)\cdot (|x|^2+|a|^2)}$
$\because |x-a|<\frac{\epsilon}{(|x|+|a|)\cdot (|x|^2+|a|^2)} \implies |x|-|a|<\frac{\epsilon}{(|x|+|a|)\cdot (|x|^2+|a|^2)}$
It feels correct to me but at the same time I have almost no confidence in my solution. (I'm trying to self-study the book after taking Calc I-III)
 A: The delta you get cannot depend on $x$. We wish to find $\delta$ such that
$$
0 < |x-a|<\delta\implies |x^4-a^4|<\epsilon
$$ 
First we do scratch work to determine a $\delta$. Note that
$$
|x^4-a^4|=|x-a||x+a||x^2+a^2|\tag{1}.
$$
We wish to bound $|x^4-a^4|$ by controlling $|x-a|$. Thus we need to find a bound for $|x+a||x^2+a^2|$ when $x$ is near $a$. To do this, take $0<|x-a|<1$ (for example). Then
$$
|x+a|\leq |x-a|+|2a|<1+2|a|\tag{2}
$$
and
$$
|x^2+a^2|\leq |x^2-a^2|+2a^2=|x-a||x+a|+2a^2<1+2|a|+2a^2\tag{3}
$$
by (2). Hence take
$$
\delta=\min\left(1,\frac{\epsilon}{(1+2|a|)(1+2|a|+2a^21+2|a|+2a^2)}\right).\tag{4}
$$
A: $$\lim_\limits{x\to a}x^4=a^4$$
Suppose an arbitrary $\varepsilon>0$ is given. We are required to find a $\delta>0$ such that
$$|x^4-a^4|<\varepsilon\Longleftarrow|x-a|<\delta$$
Let $\delta=1$
$$\begin{align}|x-a|&<1\\-1<x-a&<1\\2a-1<x+a&<2a+1\\|x+a|&<2|a|+1\end{align}$$
And we know that
$$\begin{align}|x|&=|x-a+a|\\|x|&\le|x-a|+|a|\\|x|-|a|&\le|x-a|\\|x|-|a|&\le1\\|x|^2&\le(|a|+1)^2\\x^2+a^2=|x^2+a^2|&\le(|a|+1)^2+a^2\end{align}$$
Getting back to the problem...
$$\begin{align}|x^4-a^4|&=|(x^2+a^2)(x+a)(x-a)|\\|x^4-a^4|&\le(x^2+a^2)|x+a||x-a|\\|x^4-a^4|&\le\left\{(|a|+1)^2+a^2\right\}(2|a|+1)\end{align}$$
Choosing $\delta=\min\left\{1,\dfrac{\varepsilon}{\left\{(|a|+1)^2+a^2\right\}(2|a|+1)}\right\}$ completes the proof.
A: An even easier approach is suggested by the proof of the lemma of Theorem 2, part 2, from earlier in the chapter:
If $|x-x_0| < \min(1, \frac{\epsilon}{2(|y_0|+1)})$ and $|y-y_0| < \frac{\epsilon}{2(|x_0|+1)}$, then $|xy - x_0y_0| < {\epsilon}$
In this case, you have:
$f(x)=x^4; l=a^4$
Substituting $x^2$ in for both $x$ and $y$ and $a^2$ in for both $x_0$ and $y_0$, we see that $|x^4 - a^4| = |(x^2)(x^2) - (a^2)(a^2)| < {\epsilon}$ as long as ${\delta} = \min(1, \frac{\epsilon}{2(|a^2|+1)})$
