A formal proof that the function $ x \mapsto x^{2} $ is continuous at $ x = 4 $. Problem: Show $f(x)=x^2 $ is continuous at $ x = 4$. 
That is to say, find delta such that:
$ ∀ε>0$ $ ∃δ>0 $ such that $ |x-a|<δ ⇒ |f(x)-f(a)|<ε$
Where $a=4$, $f(x)=x^2$,and $f(a)=16$. 
So in order to do to do these delta/epsilon proofs, I was originally taught to solve |f(x)-f(a)|<ε by setting it as: 
$-ε < f(x) - f(a) < ε$
Which, in this case would be: 
$-ε < x^2-16 < ε$
$16-ε < x^2 < 16+ε$
$\sqrt{16-ε} < x< \sqrt{16+ε}$
Now, since we're looking for |x-a|<δ, set $δ=\sqrt{16+ε}-4$ from the equation above. 
For the proof itself:
For $ ∀ε>0$, choose $δ=\sqrt{16+ε}-4$ and then $δ>0$ exists and implies:
$|x-a|<δ$
$ |x-4|<δ= \sqrt{16+ε}-4$
$|x-4|+4<\sqrt{16+ε}$
$|x|<\sqrt{16+ε}$
$|x^2|<16+ε$
$|x^2-16|<ε$
Q.E.D. 
The problems i'm running into are mainly with the last step. In the solutions, it uses the method of splitting up $|x^2-16|<ε$ into $|(x-4)(x+4)|<ε$ and then going from there in order to get $δ=min(1,ε/9)$. I understand how their solution works but as I was taught my way a long time ago, I'm trying to figure out why mine does/doesn't work and if it doesn't if there's a way to modify it so that it does. 
Thanks for your help!
 A: Let $\epsilon>0$ be given. Choose $\delta=\min\left \{ \epsilon/9,1 \right \}$, so if $0<\left | x-4 \right |<\delta$, then we have
$$\left | f(x)-16 \right |=\left | x-4 \right |\left | x+4 \right |<\left | x-4 \right |9<\frac{\epsilon}{9}9=\epsilon.$$
Details: I will try explain you the way I understand. Our goal is to find a $\delta>0$ such that $0<\left | x-4 \right |<\delta \implies \left | f(x)-16 \right |<\epsilon$. Since $\left | f(x)-16 \right |=\left | x-4 \right |\left | x+4 \right |$, so we directly can say $\delta=\epsilon/\left | x+4 \right |$. But we are not done at all. We have to find the other way such that $\delta$ isn't dependent on $x$. If we have a good choice of $\delta>0$, then the other smaller choice would also work. Now we can set the restriction $\delta\leq 1$, which means, that whenever we have $\left | x-4 \right |<\delta $ then we will get $\left | x-4 \right |<1$. Since $x$ is positive, we have $$\left | x-4 \right |<1\iff -3<x<5\iff 1<x+4<9\iff 1<\left | x+4 \right |<9.$$
We can finally choose $\delta=\min\left \{ \epsilon/9,1 \right \}$, since $\left | x+4 \right |<9$ makes
$$0<\left | x-4 \right |<\delta\leq\frac{\epsilon}{9}<\frac{\epsilon}{\left | x+4 \right |}.$$
A: $|x^2|<16+ε$ does not imply $|x^2-16|<ε$
For example, take $x= 0 , \epsilon = 1$.
Then $|x^2| = 0 < 17  = 16+ε$
But $|x^2-16| = |0^2-16| = 16 \geq 1 = ε $.
So in the current form, your proof is invalid. 
A: Suppose $0<|x-4|<\delta \leq 1.$ 
Then,$\ |x-4|<1 \Rightarrow |x^2-16|=|(x-4)(x+4)|<9\delta$. 
Thus you want to choose $\delta$ such that $\delta := \textbf{min}\{1,\frac{\epsilon}{9}\}$
A: I'd take a slightly different approach. Let $f: x \mapsto x^{2}$. Then 
\begin{align*}
|f(x) - f(4)| = |x^{2} - 4^{2}| \\
= | ((x - 4) + 4)^{2} - 4^{2}| \\
= | ((x - 4)^{2} + 2 (4)(x - 4) + 4^{2} ) - 4^{2}| \\
= | (x - 4)^{2} + 8(x - 4) | \\
\leq | x - 4 |^{2} + 8 | x - 4 |
\end{align*}
Now pick $\epsilon > 0$. If you can find $\delta > 0$ such that if $|x - 4| < \delta$, then $|x - 4|^{2} < \epsilon /2$, and $8|x - 4| < \epsilon / 2$, then you're set. Of course,
\begin{align*}
|x - 4|^{2} < \epsilon / 2 \iff |x - 4| < \sqrt{ \epsilon / 2 } \\
8|x - 4| < \epsilon / 2 \iff |x - 4| < \epsilon / 16,
\end{align*}
making it more evident that you can find such a $\delta > 0$ by picking it small enough that it meets both of the above conditions, e.g. $\delta < \min \{ \sqrt {\epsilon / 2}, \epsilon / 16 \}$. Admittedly, it takes a bit more writing, but in my opinion it's more intuitive, and is overall a reasonable method to go about it for polynomials.
