Finding a $δ$ for the limit $\lim_{x\to 2} x^4 = 16$ $$\lim_{x\to a} x^4 = L$$
for some arbitrary a
Picking $a$ to be 2, we get:
$$\lim_{x\to 2} x^4 = 16$$
To show that is the limit I tried doing the epsilon-delta definition of a limit to show how to find a $δ$ such that $|f(x) - L| < \epsilon $ for all x satisfying $0 < |x-a| < δ$
And here's how I attempted it:
$\forall ε>0, \exists δ>0$, such that for all x, if $0<|x-2|<δ$ then $|x^4 - 16| < ε$
$$|x^4 - 16| < ε$$
$$|(x-2)(x+2)(x^2+4)| < ε$$
$$δ: |x-2| < δ$$
I picked $δ$ to be 1, then,
$$|x-2| < 1 \Rightarrow 1 < x < 3 \Rightarrow 3 < x + 2 < 5 \Rightarrow 7 < x^2 + 4 < 9$$
so,
$$|(x-2)(x+2)(x^2+4)| < |x-2|*9 < ε \Rightarrow  |x-2| < \frac{ε}{9}$$
therefore,
$$δ: min\lbrace1, \frac{ε}{9}\rbrace$$
I was wondering if what I did was correct and if it isn't can someone show me where I might of messed up.

Given $\varepsilon>0$ we have to find a $\delta>0$ such that $|x^4-16|< \varepsilon$ whenever $0<|x-2|<\delta$. Let $0<|x-2|<\min(1,\varepsilon/500)$ ($65$ works but just for twist the argument), so $|x|<3$ (why?) and then
\begin{align}|x^4-16|= |x-2||x^3+2x^2+4x+8|\le|x-2|(|x|^3+2|x|^2+4|x|+8)\\
\le|x-2|(65)\\
<\varepsilon\end{align}
