$$\lim\_{x\to a} x^4 = L$$
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.