Prove a limit for $g(x)$ with the definition of limit only

Question

We have the function $g(x) = x^3+1$

Prove, with the definition of limit only, that the limit $L=9$ is indeed the limit of the function when $x=2$.

I started with: $\lvert g(x) - L \rvert$ : $\lvert x^3+1-9 \rvert = \lvert x^3-8 \rvert = \lvert (x-2)(x^2+2x+4) \rvert < \epsilon $

From the definition of the limit we know: $ \rvert x-2 \lvert < \delta$, so therefore: $\lvert (x-2)(x^2+2x+4) \rvert < \delta (x^2+2x+4) $

How do I proceed from here? do I need to find a condition for $ x^2+2x+4 $ ?

Thanks guys

Answer 1

$$|x^2+2x+4|=|x^2-4x+4+6x-12+12|=|(x-2)^2+6(x-2)+12| \leq |\delta^2+6\delta+12|$$

So for small $\delta$ you have (for example) $|x^2+2x+4|<13$

Answer 2

HINT: $x^2+2x+4<19$ for $1<x<3$, hence we can take $\delta=\varepsilon/19$ for $\varepsilon<1$.