# Proving the limit of a given cubic function from the epsilon-delta definition of a limit

Would anyone know how to prove the limit of this cubic equation using the epsilon delta definition?

$\lim_{x \rightarrow 2} x^3 +2x^2 -x -1 = 13$ I really don't know where to start other than inputting the values of $a$, $L$ and $f(x)$ for this example into the definition of a limit:

$0 < |x - 2| < d$ implies $|(x^3 +2x^2 -x -1) - 13| < \epsilon$

• $x^3+2x^2-x-1-13=(x-2)Q(x)\\=(x-2)(x^2+4x+7)\\$ so put at $|(x^3 +2x^2 -x -1) - 13| < \epsilon\\|(x-2)||x^2+4x+7|< \epsilon$ – Khosrotash Oct 16 '16 at 10:51

Let $f(x)=x^3+2x^2-x+1.$ Note $|f(x)-13|=|x-2||x^2+4x+7|$
For $0<|x-2|<1$, we have $1<x<3\Rightarrow12<x^2+4x+7<28$,
Let $\epsilon>0$, then take $\delta(\epsilon):=\min\{\epsilon/28,1\}$. For $0<|x-2|<\delta(\epsilon),$ we have
$$|f(x)-13|=|x-2||x^2+4x+7|<\frac{\epsilon}{28}28=\epsilon$$