Limit Proof Check: $\lim _{x \to a} x^4 = a^4$ Reviewing limits and I'm afraid I may be making mistakes, just looking for a quick proof check. 
$f(x)=x^4$, prove that $\lim _{x \rightarrow a}f(x)=a^4$ by showing how to find $\delta$ . 
This is my work. $|x^4 - a^4 | < \epsilon$ and $0<|x-a|<\delta$
Factoring: $|x^2 +a^2||x-a||x+a|< \epsilon$ next we set $|x-a|<1$ which means $-1+a<|x|<1+a$ so that 
\begin{align}
 |x^2 +a^2||x-a||x+a|& =((-1+a)^2 +a^2 )((1+a)+a)(x-a))\\
 & = |2a(1+2a^2 -2a)||x-a|\\ 
 & <\epsilon \\
\end{align}
We then have $$\delta =\min\lbrace 1, \frac{\epsilon}{ |2a(1+2a^2 -2a)|} \rbrace$$
How's it look? 
 A: Alternate Solution.
Need to find a bound for $|x^2 + a^2||x + a|$. Take for instance, $|x - a| < 1$ then
$$|x^2 + a^2||x + a| = |x^2 - a^2 + 2a^2 ||x - a + 2a| = (|x^2 - a^2 + 2a^2 |)(|x - a + 2a|)$$
This does not change the expression since $a^2 - a^2 = 0$ and $a - a = 0$
Using the triangle inequality $|x + y| \le |x| + |y|$
$$(|x^2 - a^2 + 2a^2 |)(|x - a + 2a|)$$ 
$$\le (|x^2 - a^2| + |2a^2 |)(|x - a| + |2a|)$$ 
$$\le (|x - a||x + a| + |2a^2 |)(|x - a| + |2a|)$$
$$\le (|x - a||x -a + 2a| + |2a^2 |)(|x - a| + |2a|)$$
$$\le (|x - a| \cdot (|x -a| + |2a|) + |2a^2 |)(|x - a| + |2a|)$$
$$|x - a| < 1$$ 
$$< (1 \cdot (1 + 2|a|) + 2|a^2|)(1 + 2|a|)$$
$$< (1 + 2|a| + 2|a^2|)(1 + 2|a|)$$
Therefore
$$|x^2 + a^2||x + a| < (1 + 2|a| + 2|a^2|)(1 + 2|a|)$$
Therefore
$$|x^2 + a^2||x + a||x - a| < (1 + 2|a| + 2|a^2|)(1 + 2|a|)|x - a| < \epsilon$$
$$|x - a| < \frac{\epsilon}{(1 + 2|a| + 2|a^2|)(1 + 2|a|)}$$
Take $$\delta = min \lbrace 1, \frac{\epsilon}{(1 + 2|a| + 2|a^2|)(1 + 2|a|)} \rbrace$$
A: A different factorization
works for all powers
$x^n$.
This is
$x^n-a^n
=(x-a)\sum_{k=0}^{n-1} x^ka^{n-1-k}
$.
If
$|x-a| < 1$,
then
$\begin{array}\\
|x^n-a^n|
&=|x-a||\sum_{k=0}^{n-1} x^ka^{n-1-k}|\\
&<|x-a||\sum_{k=0}^{n-1} (a+1)^ka^{n-1-k}|\\
&\le|x-a|n(|a|+1)^{n-1})\\
\end{array}
$
Therefore,
to make
$|x^n-a^n| < \delta$,
it is enough if
$|x-a|n(|a|+1)^{n-1}
 < \delta$
or
$|x-a|
< \dfrac{\delta}{n(|a|+1)^{n-1}}
$.
Inserting the restriction
$|x-a| < 1$,
this becomes
$|x-a|
< \min(1,\dfrac{\delta}{n(|a|+1)^{n-1}})
$.
