<s>[INST] Confused about the approach to this Delta Epsilon Proof I have recently started with Calculus and got into Delta Epsilon Proofs as I found it intriguing. After I went over some basic questions involving delta epsilon proofs, I have reached a question which I was not sure how to approach.
This is the questions:
Prove that  $\lim_{x\to 2} x^4 = 16$
So, what I have basically started doing is the following:
Let $δ$ = 1, and by the definition of a limit I arrived that $|x-2|$ < $1$, so I concluded that $|x + 2| < 5$. Then, I have arrived that $|x-2||x+2||x-2||x+2| < 5|x+2| * 5|x-2|$.
But I don't really know what to do next, and if my approach is even right.
 [/INST] Well, in the end we want for all $\epsilon > 0$ there is a $\delta$ based on the $\epsilon$ so that $|x-2|<\delta \implies |x^4 -16|< \epsilon$.
If we go from $|x-2| < \delta \implies..... |x^4 - 16| < h(\delta)$ for some function $h$ we can set $|x^4 - 16| < h(\delta) < \epsilon$ and solve for $\delta$ in terms of $\epsilon$.
So 
$|x-2| < \delta = h^{-1}(\epsilon)$ for some $h$ we will discover later.
$2 - \delta < x < 2 + \delta$
$(2-\delta)^4 < x < (2+\delta)^4$ (if we assume $0< \delta <2$)
$16 - 32\delta + 24\delta^2 -8\delta^3 + \delta^4 < x^4 < 16 + 32\delta + 24\delta^2 +8\delta^3 + \delta^4$
$-32\delta + 24\delta^2 - 8\delta^3 +\delta^4 < x^4 -16 <32\delta + 24\delta^2 +8\delta^3 + \delta^4$
$-32\delta - 24\delta^2 - 8\delta^3-\delta^4 <-32\delta + 24\delta^2 - 8\delta^3 +\delta^4< x^4 -16 <32\delta + 24\delta^2 +8\delta^3 + \delta^4$
(and if we assume $0 < \delta < 1$) we have
$-32\delta - 24\delta -8\delta -\delta < 32\delta - 24\delta^2 - 8\delta^3-\delta^4 <-32\delta + 24\delta^2 - 8\delta^3 +\delta^4< x^4 -16 <32\delta + 24\delta^2 +8\delta^3 + \delta^4 < 32\delta + 24\delta +8\delta + \delta$
so 
$-65\delta < x^4 -16 < 65\delta$
$|x^4 - 16 |< 65\delta$
And if we let $\delta = h(\epsilon) = \min (1, \frac {\epsilon}{65})$ then
$|x^4 - 16| < 65\delta \le \epsilon$.
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