$\lim_{x\to 1} (x^3+2x^2-2)=1$ using the definition to prove that $\lim_{x\to 1} (x^3+2x^2-2)=1$ 
let $\varepsilon>0$, and $x\in\mathbb{R}$
we must find $\delta>0$ such that if $|x-1|<\delta $ then $|x^3+2x^2-2-1|<\varepsilon$
i do :
$$
|x^3-1+2x^2-2|\leq |x^3-1|+2|x^2-1|=|x^3-1|+2 |(x-1)| |(x+1)|=
$$
$$
|x-1|(|x^2+x+1|+2|x+1|)\leq \delta (|x^2+x+1|+2|x+1|)
$$
how to continue?

$$|x^3+2x^2-2-1|= |(x-1)(x^2+3x+3)|<\delta |x^2+3x+3|<13\delta <\varepsilon $$
Provided that $$0<\delta < \min \{1,\epsilon/13\}$$
