Proving that $\lim\limits_{x \to -1}(3x^2-3)\sin(x) = 0$. 
Prove that $\lim\limits_{x \to -1}(3x^2-3)\sin(x) = 0$.

So, by the definition I have to prove that 
$$
\exists\delta>0 \text{   such that}
$$
$$
0<|x+1|<\delta \longrightarrow |(3x^2-3)\sin(x)|<\epsilon
$$
What I did:
$$
|(3x^2-3)\sin(x)|=3|x+1||x-1||\sin(x)|\leq3|x+1||x-1|
$$
Let $\delta_1=1$ then:
$$
-1<x+1<1 \longrightarrow -3<x-1<-1<3 \longrightarrow |x-1|<3
$$
So:
$$
3|x+1||x-1||\sin(x)|\leq3|x+1||x-1|<3|x+1|\cdot3<\epsilon
$$
$$
|x+1|<\frac\epsilon9
$$
So I let $\delta=\min(1,\frac\epsilon9)$ and I'm done?
Did I screw up anywhere?
What are other ways to prove this?
 A: As it is, your proof is not complete. You've made it quite likely that $\delta=\min\left\{1,\dfrac\epsilon 9\right\}$ will suffice, but it's not a formal proof yet.  To complete the formal proof, you have to show that this $\delta$ works for every $\epsilon>0$. This is quite tedious and routine, and it will be extremely similar to what you already did, however it must be done. Other than that the proof looks fine. 
N.B. I would just like to note that the definition you wrote down, is missing something quite important. It should inculde at some point the phrase for every $\epsilon>0$. It's a petty comment, I know, but it realy is quite vital to the definition.
A: The basic result 
I use here is that,
if
$\lim_{x \to a} f(x)
\ne 0$,
then
$\lim_{x \to a} f(x)g(x)
= 0$
if and only if
$\lim_{x \to a} g(x)
= 0$.
First of all,
$\sin(-1) 
\ne 0$,
so
$\lim\limits_{x \to -1}(3x^2-3)\sin(x)
 = 0$
if and only if
$\lim\limits_{x \to -1}(3x^2-3)
 = 0$.
Second,
$3x^2-3
=3(x^2-1)
=3(x+1)(x-1)
$,
and
$3((-1)-1)
=-6$,
so
$\lim\limits_{x \to -1}(3x^2-3)
 = 0$
if and only if
$\lim\limits_{x \to -1}(x+1)
 = 0$.
Finally,
you be able to prove that
$\lim\limits_{x \to -1}(x+1)
 = 0$.
