<s>[INST] Prove this limit of $x^4$ using $\epsilon - \delta$ $\lim_{x\to a} x^4 = a^4$ using epsilon and delta. 
We know $|x^4 - a^4| < \epsilon \space \text{such that} \space |x - a| < \delta$
So we have to find a $\delta$ for, which it works. 
$|x^4 - a^4| = |x^2 + a^2||x-a||x+a|$
$|x^2 + a^2||x-a||x+a| < \epsilon \space \text{such that} \space |x - a| < \delta$
Let's require, 
$|x - a| < 1 \implies |x + a| < 2|a| + 1$ 
$1 > |x - a|$
$|x - a| \ge |x| - |a|$
Therefore, 
$|x| - |a| < |x - a| < 1$
$|x| < 1 + |a|$
$x^2 < (1 + |a|)^2$
$\implies x^2 + a^2 < (1 + |a|)^2 + a^2 \implies |x^2 + a^2| < (1 + |a|)^2 + a^2$ 
$ \implies |x + a| < 2|a| + 1$ 
All together, this implies, finally:
$ |x^2 + a^2| \cdot |x+a| < ((1 + |a|)^2 + a^2)\cdot (2|a| + 1)$
This requires,  
$|x - a| < \text{min}(1, \frac{1}{((1 + |a|)^2 + a^2)\cdot (2|a| + 1)})$
Thus, $\delta = \frac{1}{((1 + |a|)^2 + a^2)\cdot (2|a| + 1)})$
Thanks, please suggest if this is correct!
 [/INST] Aside from a couple of typos, it looks pretty good! Here's a cleaned up version of your proof.

Given any $\epsilon > 0$, let $\delta = \min\left(1, \dfrac{\epsilon}{(1 + 2|a|)(1 + 2|a| + 2a^2)}\right) > 0$. Then if $0 < |x - a| < \delta$, notice that:
\begin{align*}
|x^4 - a^4|
&= |x - a| \cdot |x + a| \cdot |x^2 + a^2| \\
&= |x - a| \cdot |(x - a) + 2a| \cdot |(x - a)^2 + 2a(x - a) + 2a^2| \\
&\leq |x - a| \cdot (|x - a| + 2|a|) \cdot (|x - a|^2 + 2|a||x - a| + 2a^2) &\text{triangle inequality}\\
&< |x - a| \cdot (1 + 2|a|) \cdot (1 + 2|a| + 2a^2) &\text{} |x - a| < \delta \leq 1 \\
&< \frac{\epsilon}{(1 + 2|a|)(1 + 2|a| + 2a^2)} \cdot (1 + 2|a|)(1 + 2|a| + 2a^2)
&(\star)\\
&= \epsilon
\end{align*}
where the second-to-last step $(\star)$ follows because:
$$
|x - a| < \delta \leq \frac{\epsilon}{(1 + 2|a|)(1 + 2|a| + 2a^2)}
$$
</s>