# Prove that $\lim_{x \to 2} x^3 = 8$ by using epsilon-delta

Prove that $$\lim_{x \to 2} x^3 = 8$$

My attempt,

Given $\epsilon>0$, $\exists \space \delta>0$ such that if $$|x^3-8|<\epsilon \space \text{if} \space 0<|x-2|<\delta$$

$$|(x-2)(x^2+2x+4)|<\epsilon$$

I'm stuck here. Hope someone could continue the solution and explain it for me. Thanks in advance.

$$x^2 + 2x + 4 = (x-2)^2 + 6(x-2) + 12$$
For $\varepsilon > 0$ you would take $\delta < \min\left\{\frac\varepsilon{19}, 1\right\}$ because then $|x-2| < \delta$ implies:
$$|x^3 - 8| \le \delta(\delta^2 + 6\delta + 12) < \frac\varepsilon{19} (1 + 6 + 12) = \varepsilon$$
For $x\in [1,3]\quad |(x-2)(x^2+2x+4)|\le 19|x-2|<\epsilon$