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

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

HINT: $x^2+2x+4<19$ for $1<x<3$, hence we can take $\delta=\varepsilon/19$ for $\varepsilon<1$.
• How did you know that $x2+2x+4 < 19$ for $1 < x < 3$ ? – FigureItOut Nov 14 '14 at 20:46
• @user1326293 It is increasing for positive $x$, and value in 3 is equal to 19. – Przemysław Scherwentke Nov 14 '14 at 20:48
• This is not a good answer. You should specify that you are assuming a priori that $\delta < 1$ so that you can have $|x-2| < 1$. At the end, you take the min of $1$ and $\epsilon/19$. – user139708 Nov 14 '14 at 20:49
$$|x^2+2x+4|=|x^2-4x+4+6x-12+12|=|(x-2)^2+6(x-2)+12| \leq |\delta^2+6\delta+12|$$
So for small $\delta$ you have (for example) $|x^2+2x+4|<13$