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?
On
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| $.
On
You can go the other way
$|x - 1|<\delta$
$-\delta < x-1 <\delta$
$1-\delta < x < 1+\delta$
If we assume $0 < \delta < 1$ then
$(1-\delta)^3 + 2(1-\delta)^2 - 2 < x^3 + 2x^2-2 < (1+\delta)^3 + 2(1+\delta)^2 - 2$
$-\delta^3 +4\delta^2 -5\delta + 1 < x^3 + 2x^2-2<\delta^3 + 4\delta^2 +5\delta + 1$
Now if we assume $0 < \delta<1$ then $0< \delta^3 <\delta^2 < \delta < 1$ so
$-\delta^3 + 4\delta^2 - 5\delta + 1 > -\delta^3 + 4\delta^3 -5\delta + 1=$
$-3\delta^3 - 5\delta + 1 > -3\delta - 5\delta + 1 = -8\delta + 1$.
And $\delta^3 + 4\delta^2 +5\delta + 1 < \delta + 4 \delta + 5 \delta + 1=10\delta + 1$.
So $-10\delta + 1<-8\delta+1 < x^3 + 2x^2-2< 10\delta + 1$
So $-10\delta < x^3 + 2x^2 -2 - 1 < 10 \delta$
$|(x^3 + 2x^2 - 2)-1|< 10\delta$.
So if we set $\delta = \min(\frac \epsilon {10}, 1)$ then
$|x-1| < \delta \implies |(x^3 + 2x^2 - 2)-1|< 10\delta \le \epsilon$
$$|x^3+2x^2-2-1|= |(x-1)(x^2+3x+3)|<\delta |x^2+3x+3|<13\delta <\varepsilon $$
Provided that $$0<\delta < \min \{1,\epsilon/13\}$$