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.
 A: Let $\epsilon > 0$. Then you want to find $\delta$ dependent on $\epsilon$ such that that for any $x$ in $0<|x-2|<\delta$, we have $|x^4-16|<\epsilon$. Now notice that for $x$ in this range,
\begin{align*}
|x^4-16| &= |x+2||x-2||x^2+4|\\
&\leqslant (|x-2|+4)\,|x-2|\,(|x-2|^2+4|x-2|+8)\quad \text{(triangle inequality)}\\
&<\delta^4+8\delta^3+24\delta^2+32\delta.
\end{align*}
Now if some $\delta > 1$ makes this less than $\epsilon$, it's obvious that any $\delta < 1$ will also work.  Thus we can assume that $\delta < 1$, which would imply
\begin{align*}
|x^4-16| &<\delta^4+8\delta^3+24\delta^2+32\delta\\
&< \delta+8\delta+24\delta+32\delta \quad\text{(since $\delta<1$, higher powers are smaller)}\\
&=65\delta < \epsilon,
\end{align*}
if we pick $\delta=\frac\epsilon{65}$, completing the proof.
Thus, you need to take $\boxed{\delta=\min\{1,\tfrac\epsilon{65}\}}$.
A: No, the proof must show that for any $\epsilon$ is it possible to find a $\delta$ such that… (see below). So you assume that $\epsilon$ is given and you must show that a suitable $\delta$ exists. Usually you do it constructively, i.e. exhibit a formula $\delta=f(\epsilon)$ that works.
Here the condition to be reached is
$$|x-2|<\delta\implies|x^4-16|<\epsilon.$$
Notice that
$$|x^4-16|=|x-2||x^3+2x^2+4x+8|.$$
Then if we restrict $\delta$ to be $<1$ (i.e. $x$ in $(1,3)$), we certainly have
$$|x^3+2x^2+4x+8|>15$$ and
$$|x-2|<\frac\epsilon{15}.$$
So $$\delta=\min\left(1,\frac\epsilon{15}\right)$$ is a function that works.
A: 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
$|x^4 - 16| < 65\delta \le \epsilon$.
