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.

Let's start at d = 1 ( d = delta ) : 1 < x < 3 ==> 1 < x^2 < 9 ==> 5 < x^2 + 4 < 13 ( yours is 9 ). Next, 3 < x + 2 < 5 ==> /x + 2/ < 5. Finally: /x^2 - 16/ = /(x - 2)(x + 2)(x^2 + 4)/ < 5*13*/x - 2/ = 65*/x - 2/. We need that 65*/x - 2/ < e ( e = epsilon ) ==> /x - 2/ < e/65. We want that /x - 2/ < 1 and also that /x - 2/ < e/65. So we simply pick d = min{1, e/65} > 0, then we should be done.
