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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.

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Note that

$$x^2 + 2x + 4 = (x-2)^2 + 6(x-2) + 12$$

so \begin{align}|(x-2)(x^2 + 2x + 4)| &= |(x-2)((x-2)^2 + 6(x-2) + 12)|\\ &\le |x-2|(|x-2|^2 + 6|x-2| + 12) < \delta(\delta^2 + 6 \delta + 12) \end{align}

For $\varepsilon > 0$ you would take $\delta < \min\left\{\frac\varepsilon{19}, 1\right\}$ because then $|x-2| < \delta$ implies:

$$|x^3 - 8| \le \delta(\delta^2 + 6\delta + 12) < \frac\varepsilon{19} (1 + 6 + 12) = \varepsilon$$

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For $x\in [1,3]\quad |(x-2)(x^2+2x+4)|\le 19|x-2|<\epsilon$

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