$\lim_{x\to 1} (x^3+2x^2-2)=1$ 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?

To make it easier
(I always prefer to have variables
go to zero),
let
$x = y+1$.
Then
$
f(x)=x^3+2x^2-2
=(y+1)^3+2(y+1)^2-2\\
=y^3+3y^2+3y+1
+2(y^2+2y+1)
-2\\
=y^3+5y^2+7y\\
=y(y^2+5y+7)
$
and it's easy to see
what happens as
$y \to 0$.
In particular,
if
$|y| < 1$
(i.e., $|x-1| < 1$),
then
$|f(x)|
\lt 13|y|
= 13|x-1|
$.
